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Conductor  per  Mile. 


TRANSMISSION  LINE  FORMULAS 


FOR 


ELECTRICAL  ENGINEERS 


AND 


ENGINEERING  STUDENTS 


BY 


HERBERT    BRISTOL    DWIGHT,  B.  Sc. 

ASSOCIATE   OF    THE    AMERICAN    INSTITUTE   OF 
ELECTRICAL    ENGINEERS 


NEW    YORK 

D.   VAN   NOSTRAND   COMPANY 

25  PARK  PLACE 

1913 


Engineering 
Library 


V 


COPYRIGHT,  1913,  BY 
D.  VAN   NOSTRAND  COMPANY 


Stanbopc  jpress 

F.    H.  GILSON  COMPANY 
BOSTON,  U.S.A. 


PREFACE. 


THE  object  of  this  book  is  to  compile  a  set  of  instructions 
for  engineers,  which  will  enable  them  to  make  electrical 
calculations  for  transmission  lines  with  the  least  possible 
amount  of  work. 

The  chart  and  working  formulas  have  for  the  most  part 
been  developed  independently  by  the  author.  Where  the 
same  or  similar  methods  have  been  previously  published, 
the  fact  is  generally  stated  in  the  footnotes,  but  it  has  not 
been  found  possible  to  make  these  references  absolutely 
complete. 

The  second  part  of  the  book  is  for  reference  and  contains 
the  derivation  of  the  principal  formulas  used  in  connec- 
tion with  transmission  lines.  As  many  recent  articles  on 
transmission  lines  make  use  of  formulas  which  are  only 
roughly  approximate,  or  are  even  incorrect,  a  reliable  col- 
lection of  formulas,  with  the  method  of  obtaining  them, 
should  be  found  valuable. 

It  should  not  be  presumed,  because  the  second  part  of 
the  book  requires  the  use  of  the  integral  calculus,  that 
the  working  formulas  will  require  a  knowledge  of  higher 
mathematics.  The  first  five  or  six  chapters  are  complete 
in  themselves,  and  are  planned  for  the  use  of  those  who 
have  an  ordinary  acquaintance  with  alternating-current 
calculations. 

H.  B.  DWIGHT. 

HAMILTON,  CANADA, 
August,  i pi 2. 

in 

266994 


CONTENTS. 


PART  I. —  WORKING  FORMULAS. 

CHAPTER  PAGE 
Frontispiece  —  REGULATION  CHART 

PREFACE iii 

I.  INTRODUCTION i 

II.   ELEMENTS  or  A  TRANSMISSION  LINE 4 

III.  REGULATION  CHART 8 

IV.  FORMULAS  FOR  SHORT  LINES 16 

V.  K   FORMULAS 25 

VI.  CONVERGENT  SERIES 41 

PART  II.  — THEORY. 

VII.  CONDUCTORS 53 

VIII.  TRANSMISSION  LINE  PROBLEMS 57 

IX.  REACTANCE  OF  WIRE,  SINGLE-PHASE 62 

X.  SKIN  EFFECT 67 

XI.  REACTANCE  OF  CABLE,  SINGLE-PHASE 76 

XII.  REACTANCE  OF  TWO-PHASE  AND  THREE-PHASE  LINES 79 

XIII.  CAPACITY  OF  SINGLE-PHASE  LINE 84 

XIV.  CAPACITY  OF  TWO-PHASE  AND  THREE-PHASE  LINES 93 

XV.  THEORY  OF  CONVERGENT  SERIES 98 

PART  III.  — TABLES. 
TABLE 

REGULATION  CHART 

I.  FORMULAS  FOR  SHORT  LINES 104 

Conditions  given  at  Receiver  End. 

II.  FORMULAS  FOR  SHORT  LINES 106 

Conditions  given  at  Supply  End. 

III.  K  FORMULAS 108 

Conditions  given  at  Receiver  End. 

IV.  K  FORMULAS 112 

Conditions  given  at  Supply  End. 
V 


VI  CONTENTS 

TABLE  PAGE 

V.  CONVERGENT  SERIES 1 16 

Conditions  given  at  Receiver  End. 

VI.  CONVERGENT  SERIES 117 

Conditions  given  at  Supply  End. 

VII.  RESISTANCE  OF  COPPER  WIRE  AND  CABLE 118 

VIII.  RESISTANCE  OF  ALUMINUM  CABLE 120 

IX.  REACTANCE  OF  WIRE,   25  CYCLES 121 

X.  REACTANCE  OF  CABLE,  25  CYCLES 122 

XI.  REACTANCE  OF  WIRE,  60  CYCLES 124 

XII.  REACTANCE  OF  CABLE,  60  CYCLES 125 

XIII.  CAPACITY  SUSCEPTANCE  OF  WIRE,  25  CYCLES 127 

XIV.  CAPACITY  SUSCEPTANCE  OF  CABLE,  25  CYCLES 128 

XV.  CAPACITY  SUSCEPTANCE  OF  WIRE,  60  CYCLES 130 

XVI.  CAPACITY  SUSCEPTANCE  OF  CABLE,  60  CYCLES 131 

XVII.  POWER  FACTOR  TABLE 133 

ALPHABETICAL  INDEX 135 


Transmission  Line  Formulas 


PART    I. 
WORKING    FORMULAS. 


CHAPTER  I. 

INTRODUCTION. 

THE  determination  of  the  electrical  characteristics  of 
transmission  lines  is  a  problem  of  considerable  practical 
importance  to  engineers.  It  occurs  frequently  in  electri- 
cal engineering  work,  and  various  methods  have  been  pro- 
posed for  carrying  out  the  calculations  with  greater  or  less 
degrees  of  accuracy.  Unfortunately,  most  of  the  methods 
so  far  presented  have  been  such  as  to  require  special  mathe- 
matical skill  in  using  unfamiliar  forms  such  as  hyperbolic 
sines,  etc.  Even  the  approximate  methods,  whose  results 
are  not  intended  to  be  reliable  for  lines  of  considerable 
length,  are  often  too  cumbersome  to  be  used  by  an  engi- 
neer who  has  not  time  to  make  himself  thoroughly  familiar 
with  them.  The  result  has  been  that  engineers  have  often 
been  satisfied  with  calculations  for  practical  cases  which 
were  not  nearly  as  correct  as  they  might  easily  have  been. 

Working  methods  have  been  developed,  and  are  presented 
in  the  following  chapters,  for  solving  problems  in  connec- 
tion with  actual  transmission  lines.  As  these  working 
methods  involve  comparatively  simple  operations  in  algebra 


2        ,  TRANSMISSION  .LINE  FORMULAS 

.>\         .,,-  -i  {..*;/.\ 

and  arithmetic  only,  they  should  be  found  useful  by  all 
electrical  engineering  students  and  engineers.  Groups  of 
problems  with  answers  are  added  to  provide  practice  in  the 
use  of  the  formulas.  The  working  formulas  can  be  imme- 
diately used  by  electrical  engineers  without  the  delay 
caused  by  working  out  the  true  meaning  and  correct  oper- 
ation of  long  and  intricate  systems  of  calculation.  They 
are  also  arranged  to  require  the  minimum  amount  of  labor 
for  routine  work  where  many  lines  are  to  be  calculated. 

The  first  method,  which  is  described  in  Chapter  III,  is  in 
the  form  of  a  chart  which  gives  the  regulation  or  voltage 
drop  of  a  line,  and  which  also  shows  directly  the  required 
size  of  conductor  for  given  conditions. 

In  Chapter  IV  are  given  formulas  for  distribution  lines 
and  transmission  lines  only  a  few  miles  long.  These  are 
extended  in  Chapter  V  by  means  of  the  constant  K  to 
apply  to  transmission  lines  of  any  length  in  ordinary 
practice. 

For  purposes  of  checking  different  formulas,  and  for 
the  calculation  of  extremely  long  and  unusual  lines,  the 
fundamental  formulas  of  transmission  lines  are  expressed 
by  rapidly  converging  series  in  Chapter  VI.  While  these 
series  require  much  more  arithmetical  work  than  the  K 
formulas,  they  will  give  the  exact  results  to  any  degree  of 
accuracy  desired.  The  method  of  convergent  series  in- 
volves the  use  of  complex  numbers,  that  is,  numbers  in 
which  "j"  terms  appear.  They  are  easier  to  handle, 
however,  than  logarithms,  sines  and  cosines  of  angles,  or 
hyperbolic  functions,  and  therefore  the  use  of  these  other 
mathematical  functions  has  been  avoided. 

Each  of  the  above  groups  of  working  formulas  is  printed 
in  a  table,  ready  for  practical  use.  The  tables  will  be 


INTRODUCTION  3 

found  in  the  collection  at  the  back  of  the  book,  as  well  as 
in  the  separate  chapters  describing  them. 

When  any  formula  is  given  which  uses  approximations, 
the  limits  of  its  accuracy  should  be  clearly  stated  so  that 
one  can  tell  at  a  glance  whether  the  method  is  sufficiently 
accurate  for  the  purpose  in  hand,  or  whether  a  longer 
method  giving  greater  accuracy  is  desirable.  This  is  espe- 
cially necessary  in  the  calculation  of  transmission  lines, 
because  approximate  formulas  are  quite  permissible  for 
lines  only  a  few  miles  long,  but  become  very  untrust- 
worthy when  the  length  is  increased  to  one  hundred  miles 
or  more.  For  this  reason,  each  table  of  formulas  has  its 
percentage  and  range  of  accuracy  printed  in  a  prominent 
position,  so  that  the  most  suitable  method  for  any  case 
may  be  instantly  chosen. 


CHAPTER  II. 
ELEMENTS  OF  A  TRANSMISSION  LINE. 

THE  essential  elements  of  a  transmission  line  have  been 
described  many  times,  but  a  short  discussion  of  them, 
with  an  explanation  of  some  of  the  terms  used  in  connec- 
tion with  the  subject,  may  be  useful  before  proceeding 
with  the  actual  calculations. 

A  transmission  line  consists  of  two  or  more  conductors 
insulated  from  each  other  so  that  they  can  carry  energy  by 
electric  currents  to  some  more  or  less  distant  point. 

The  conductors  may  be  solid  copper  wires,  copper  cables, 
or  aluminum  cables.  The  diameters  and  resistances  of 
various  standard  conductors  are  given  in  Tables  VII  and 
VIII,  pages  118  to  120.  It  will  be  noted  that  the  exact 
value  of  the  resistance  of  a  conductor  differs  slightly  when 
a  direct  current,  and  an  alternating  current  of  25  or  60 
cycles,  is  flowing.  This  is  due  to  the  "skin  effect,"  by 
which  an  alternating  current  tends  to  flow  near  the  surface 
of  a  conductor,  as  explained  in  Chapter  X.  The  drop  in 
voltage  due  to  resistance  is  proportional  to  the  current 
and  is  in  phase  with  it  when  the  current  is  alternating. 

Only  overhead  lines,  carrying  alternating  currents,  will 
be  considered  in  this  book.  Such  lines  are  supported  by 
poles  or  steel  towers  at  a  considerable  height  above  the 
ground.  The  conductors  are  separated  from  each  other  by 
a  distance  which  may  be  several  inches  or  several  feet. 
This  distance  is  called  the  " spacing"  of  the  conductors 

4 


ELEMENTS  OF  A  TRANSMISSION  LINE  5 

and  it  has  an  important  bearing  on  the  electrical  charac- 
teristics of  the  line. 

An  alternating  magnetic  field  is  formed  around,  and  in- 
side of,  conductors  carrying  alternating  currents.  This 
field  generates  a  voltage  along  the  conductors  which  is 
proportional  to  the  current,  like  the  voltage  drop  due  to 
resistance,  but  which  is  90°  out  of  phase  with  the  current. 
This  voltage  is  called  the  reactance  drop.  Tables  of  re- 
actance of  transmission  lines  will  be  found  in  Part  III. 

Since  the  voltage  drop  in  a  transmission  line  is  due  to 
resistance  and  reactance,  a  simple  line  may  be  considered 
to  be  made  up  of  the  elements  shown  in  Fig.  i.  If  R  is 

Resistance 


Receiv- 

Suppty  Eg  E    er0r 

Resistance  ^^Sr«          i      Load. 

Fig.  i. 

the  total  resistance,  the  voltage  drop  in  phase  with  the 
current  /  will  be  IR,  and  if  X  is  the  total  reactance,  the 
voltage  drop  in  quad- 
rature with  the  current 
will  be  IX. 

The  vector  diagram  of 
the  above  quantities  will 
be  as  in  Fig.  2.  The  cur- 
rent is  in  general  not  in 
phase  with  the  voltage  Fig- 2- 

E,  but  lags  behind  it  by  an  angle  6,  according  to  the 
power  factor,  cos  6,  of  the  load.  The  resistance  drop  IR 
will  therefore  not  be  added  directly  to  E,  but  must  be 
added  vectorially,  along  with  the  reactance  drop  IX,  as  in 


6  TRANSMISSION  LINE  FORMULAS 

Fig.  2.  It  is  evident  that  the  voltages  E  and  E8,  and  the 
power  factors  cos  6  and  cos  0,  at  the  two  ends  of  the  line, 
are  not  the  same  in  value. 

A  long  transmission  line  acts  as  a  condenser  and  this 
fact  also  must  be  taken  into  account.  A  condenser  con- 
sists of  two  electrical  conductors  placed  close  together  but 
insulated  from  each  other  so  that  a  direct  current  cannot 
pass  between  them.  However,  if  an  alternating  voltage 
be  applied  between  them,  a  charge  of  electricity  propor- 
tional to  the  electrostatic  capacity  of  the  condenser  will 
flow  into  and  out  of  the  conductors.  The  result  is  that  an 
alternating  current  will  appear  to  flow  between  them,  pro- 
portional to  the  t  capacity  susceptance  of  the  condenser. 
This  current,  called  the  charging  current,  will  be  90° 
out  of  phase  with  the  voltage,  and,  unlike  most  currents 
in  ordinary  practice,  it  will  lead  the  voltage  in  phase,  in- 
stead of  lagging  behind  it.  The  amount  of  the  charging 
current  may  be  determined  by  means  of  the  tables  of 
capacity  susceptance  of  transmission  lines,  in  Part  III. 

A  current  in  phase  with  the  voltage  will  flow  between 
the  conductors,  but  it  is  only  noticeable  at  very  high 
voltages.  Part  of  it  is  a  leakage  current  flowing  over  the 
insulators,  and  part  is  a  discharge  through  the  air,  and 
produces  the  glow  called  corona,  on  high-voltage  con- 
ductors. 

The  elements  of  a  transmission  line  accounting  for  the 
leakage  current  and  charging  current  are  shown  in  Fig.  3, 
in  which  resistances  and  condensers  are  shunted  across  the 
line  all  along  its  length. 

Considering  for  the  present  that  the  voltage  of  the  line 
is  the  same  at  all  parts  and  is  equal  to  £,  the  current  in 
phase  with  E  flowing  across  from  one  conductor  to  tjie 


ELEMENTS   OF  A  TRANSMISSION  LINE 


other  will  be  EG,  where  G  is  the  total  conductance  between 
the  wires.     So  also,  if  B  is  the  capacity  susceptance  of  the 


Supply 


P  Receiv- 
E   er. 


Fig.  3. 

line  considered  as  a  condenser,  EB  will  be  the  value  of  the 
shunted  current  in  quadrature 
with  E. 

The  vector  diagram  for  the 
line  indicated  in  Fig.  3  (neg- 
lecting the  voltage  drop  in  the 
conductors)  is  shown  in  Fig.  4. 
It  is  seen  that  the  current  I8  at 
the  supply  end  is  different  in 
magnitude  and  phase  from  the  current  /  at  the  receiver. 

In  order  to  calculate  the  combined  effect  of  the  above 
phenomena,  formulas  must  be  used  which  will  take  into 
account  the  fact  that  the  resistance,  capacity,  etc.,  are 
uniformly  distributed  along  the  line,  and  that  the  line  cur- 
rent and  voltage  are  different  at  all  parts  of  the  line. 


Fig.  4. 


CHAPTER  III. 
REGULATION  CHART. 

THE  characteristic  of  a  transmission  line  which  limits 
the  load  it  may  carry  is  its  regulation,  or  the  variation  in 
voltage  which  occurs  when  the  load  is  thrown  on  and  off. 
This  is  especially  true  when  the  load  has  a  low  power 
factor,  which  is  the  case  in  most  instances  at  the  present 
time. 

For  estimating  the  regulation  of  a  line,  or  the  size  of 
conductor  required,  the  regulation  chart  which  forms  the 
frontispiece  of  the  book  may  be  used,  and  it  will  give  the 
required  result  much  quicker  than  any  method  of  calcula- 
tion. An  extra  copy  of  the  chart  is  inserted  at  the  back 
of  the  book;  it  may  be  found  useful  for  cutting  out  and 
mounting  on  cardboard. 

The  chart  is  accurate  within  approximately  \  of  i%  of 
full  line  voltage,  when  the  regulation  is  less  than  10%  and 
the  line  is  not  more  than  100  miles  long. 

In  using  the  chart,*  one  places  a  straightedge  across  it 
from  the  point  on  the  left  corresponding  to  the  spacing  of 
the  transmission  line,  to  the  point  on  the  right,  correspond- 
ing to  the  resistance  of  the  conductor  per  mile.  The  reg- 
ulation voltage,  V,  per  total  ampere  per  mile  of  line,  is 
then  read  directly  from  the  chart  for  the  power  factor  of 
load  considered.  The  regulation  is  taken  as  the  change  in 

*  The  process  of  using  the  chart  is  similar  to  that  used  with  the  trans- 
former regulation  and  efficiency  charts  published  by  J.  F.  Peters,  Electric 
Journal,  December,  1911. 

8 


REGULATION  CHART  9 

load  voltage  when  the  load  is  thrown  on  or  off,  assuming 
constant  supply  voltage. 

The  total  regulation  is  quickly  figured  on  the  slide  rule 
from  the  following  formula  for  two-phase  (four  wire)  or 
three-phase  lines: 

™      i  .•      w  u        looo  K.V.A.  X  IV 
Regulation  Volts  =  - 

where  K.V.A.  =  Kilovolt-amperes  of  load,  at  the  receiver 

end. 

E  =  Line  voltage  at  the  load,  or  receiver  end. 
/  =  Length  of  line  in  miles. 

For  single-phase  lines  use  2V  instead  of  F,  making  the 
formula  as  follows: 

_>      .    .      _,  u        looo  K.V.A.  X  I  X  2  F 
Regulation  Volts  =  - 

The  regulation  volts  may  be  expressed  as  a  percentage 
of  E  to  give  the  per  cent  regulation,  and  a  formula  is  given 
on  the  chart  for  obtaining  this  result  directly. 

The  line  drop,  or  difference  in  voltage  between  the  supply 
end  and  the  receiver  end  of  the  line,  is  the  same  as  the 
regulation  for  lines  less  than  about  20  miles  long,  but  for 
longer  lines  the  effect  of  the  charging  current  must  be  taken 
into  account  by  the  formula 

Line  Drop  =  Regulation  Volts  — 


aooo/ 

where  K  =  2.16  for  60  cycles, 

and  K  =  .375  for  25  cycles. 

It  is  seen  that  the  voltage  due  to  the  charging  current 
is  proportional  to  the  line  voltage  E,  and  to  the  square 
of  the  number  of  miles,  but  is  independent  of  the  size  or 
spacing  of  the  conductors,  within  the  assigned  limit  of  ac- 
curacy. The  constant  K  does  not  need  to  be  used  in  the 


10  TRANSMISSION  LINE   FORMULAS 

v 

formula  for  regulation,  since  the  charging  current  is  present 
at  both  no  load  and  full  load. 

In  selecting  the  spacing  point  on  the  chart,  one  notes 
whether  the  frequency  is  25  or  60  cycles,  and  whether  the 
conductor  is  of  copper  or  aluminum.  The  spacing  points 
are  the  same  for  both  solid  wire  and  cable.  When  the 
wires  of  a  three-phase  line  are  not  spaced  at  the  corners  of 
an  equilateral  triangle,  but  are  at  irregular  distances  a,  b, 
and  c  from  each  other,  as  in  Figs.  5  and  6,  the  equivalent 
spacing  _ 

5  =  v  abc 
should  be  used. 


...........  c  .........  ~ 

Fig.  6.    Irregular  Flat  Spacing. 

K—  --  a  .....  -><-  ......  a  ----->j 

I  ............  2a  ............  j 

Fig.  5~    Irregular  Triangular  Spacing.  Fig.  7.    Regular  Flat  Spacing. 

With  regular  flat  spacing,  as  in  Figs.  7  and  8,  the  equa- 
tion for  the  equivalent  spacing  becomes  simply 

s  =  i.  26  a. 

It  makes  no  difference  whether  the  plane  of  the  wires 
with  flat  spacing  is  horizontal,  vertical  or  inclined. 

The  spacing  of  a  two-phase  line  is  the  average  distance 
between  wires  of  the  same  phase.  The  distance  between 
wires  of  different  phases  is  not  considered. 

The  points  marked  on  the  resistance  scale  at  the  right 
of  the  chart  are  for  cables  at  20°  C.,  assuming  hard-drawn 
copper  of  a  conductivity  equal  to  97.3%  of  the  Annealed 


REGULATION  CHART  II 

Copper  Standard,  and  hard-drawn  aluminum  of  60.86% 
conductivity,  and  allowing  an  increase  of  i% 
in  resistance  for  the  effect  of  spiralling  of  the 
wires  in  the  cable.  However,  these  resistance 
points  are  placed  on  the  chart  for  convenience 
only,  and  are  not  essential.  If  other  assump- 
tions are  made,  or  if  other  sizes  of  conductor 
are  used,  all  that  is  needed  is  to  find  the  re- 
sistance of  the  conductor  per  mile,  and  use  the 
corresponding  resistance  point  on  the  chart  Fis-8-  Regular 
to  find  "V."  FtatSpactag- 

One  of  the  most  common  problems  in  estimating  new 
projects  is  to  determine  the  size  of  wire  needed  for  any 
given  value  of  regulation,  and  the  chart  will  be  found 
especially  applicable  to  this  work.  "V"  is  first  found  from 
the  equation, 

v  =      %  Reg'n  X  E2 
~  100,000  K.V.A.  X  I ' 

Then  lay  a  straightedge  through  "V"  and  the  point  for 
the  spacing  to  be  used,  and  the  nearest  size  of  conductor 
can  be  seen,  at  a  glance,  on  the  resistance  scale  at  the  right. 
The  chart  is  quite  as  useful  for  finding  the  voltage  drop, 
or  required  size  of  conductor,  for  distribution  lines  a  few 
hundred  feet  long  as  it  is  for  transmission  lines  many  miles 

long. 

PROBLEM  A. 

Find,  by  means  of  the  chart,  the  regulation  and  line  drop  for  the 
following  set  of  conditions: 

Length  of  line 100  miles. 

Spacing 8  feet. 

Conductor No.  3  copper  cable. 

Load  (measured  at  receiver  end) ,  3000  K. V. A., 
66,000  volts,  90%  P.F.,  three  phase,  60  cycles. 


12  TRANSMISSION  LINE  FORMULAS 

Lay  a  straightedge  from  the  8-foot  spacing  point  (60  cycles,  copper 
conductor)  to  the  point  on  the  resistance  scale  for  No.  3  copper 
cable.  It  is  found  to  cross  the  90%  P.F.  line  at  the  reading  1.344. 
Then,  by  the  formula  on  the  chart, 

^  T.       ,  ,.  100,000  X  3000  x  100  x  1.344 

Per  cent  Regulation  =  z — — — 223 

66,000  X  66,000 

=  9.26%,  or  approximately  9.3%. 

The  calculated  value  of  the  regulation  of  this  line  is  9.40%  (Chap. 
VI,  Prob.  2),  so  that  the  error  involved  in  using  the  chart  is  less  than 
I  of  i%  of  line  voltage. 

The  per  cent  line  drop,  according  to  the  chart,  is 
9.26  —  100  x  2.16  x  T£<T  =  7-io%. 

As  the  calculated  value  is  7.08%  (Chap.  VI,  Prob.  2),  the  error 
from  the  chart  is  less  than  rV  of  i%  of  the  line  voltage. 

PROBLEM  B. 

1  To  find  the  size  of  copper  required  to  give  approximately  10% 
voltage  drop  in  the  following  case: 

Length  of  line 3  miles. 

Flat  spacing  as  in  Fig.  7.    Wires  2  feet  apart. 
Load  (measured  at  receiver  end),  250  K.V.A., 
2200  volts,  85%  P.F.,  three  phase,  60  cycles. 
First,  find  V  from  the  formula  on  the  chart. 

V  =   10  X  2200  X  2200    = 

100,000  X  250  X  3 

The  equivalent  spacing  is  1.26  X2,  or  2.52  feet.  The  proper 
spacing  point  will  therefore  be  just  below  the  spacing  point  for  2^ 
feet,  copper  conductor,  60  cycles.  Lay  a  straightedge  from  this 
point  to  the  reading  0.64  on  the  line  for  85%  P.F.  and  it  cuts  the 
resistance  scale  at  0.36  ohm  per  mile.  The  nearest  size  of  copper  is 

seen  to  be  No.  ooo. 

PROBLEM  C. 

Find  the  voltage  drop  of  the  following  two-phase  line: 

Length  of  line 80  miles. 

Spacing 10  feet. 

Conductor No.  oo  aluminum  cable. 

Load  (measured  at  receiver  end),  15,000  K.V.A., 
100,000  volts,  95%  P.F.,  two  phase,  25  cycles. 


REGULATION  CHART  13 

Laying  a  straightedge  across  the  chart  from  the  lo-foot  spacing 
point  for  25  cycles  and  aluminum  conductor,  to  the  resistance  point 
for  No.  oo  aluminum,  the  value  of  V  for  95%  P.F.  is  found  to  be 
0.750.  Then  the  line  drop,  in  volts,  is  equal  to 

1000x15.000x80x0.750  _      o          o  o8   o  o8 

100,000 

=  9000  —  240 
=  8760  volts. 

The  calculated  value  is  8810  volts  (Chap.  VI,  Prob.  i).  The 
error  from  the  chart  is  0.05%  of  line  voltage.  . 

PROBLEM  D. 

Find  the  regulation  of  the  following  single-phase  line: 

Length  of  line 15  miles. 

Spacing 3  feet. 

Conductor No.  o  copper  wire. 

Load  (at  receiver  end),  300  K.V.A.,  50%  P.F., 

11,000  volts,  single  phase,  60  cycles. 
From  the  chart,    V  =  0.851. 
Therefore  Regulation  =  i°o.°oo  X«oo  X  iS  X  a  Xo.8Si  _  6      % 

11,000  X  11,000 

[Calculated  value,  6.40%  (Chap.  IV,  Prob.  A).  Error  from  chart, 
0.07%  of  line  voltage.] 

PROBLEM  E. 

Find  the  K.V.A.  which  can  be  delivered  at  the  end  of  the  following 
line,  with  8%  regulation: 

Length  of  line 75  miles. 

Spacing 8  feet,  regular  flat  spacing. 

Conductor No.  oo  aluminum  cable. 

Character  of  load  (at  receiver  end), 

88,000  volts,  85%  P.F.,  three  phase,  25  cycles. 
Equivalent  spacing  5  =  8  X  1.26  =  10.08  feet. 

V  =  0.755. 

KVA  =  8  X  88,000  X  88,000 
100,000  X  0.755  X  75 
=  10,900. 


14  TRANSMISSION  LINE   FORMULAS 

PROBLEMS,  CHAPTER  III. 

(REGULATION  CHART.) 

1.  Find  the  size  of  copper  cable  which  is  needed  to  deliver  200 
K.V.A.  at  a  distance  of  3  miles  with  10%  drop  or  less. 

Spacing  of  line 2  feet. 

Character  of  load    (at  receiver   end),    2200  volts, 
80%  P.F.,  three  phase,  60  cycles. 

[Ans.   No.  o.] 

2.  Assuming  No.  o  copper  cable  for  the  previous  problem,  find 
the  volts  drop  in  the  line. 

[Ans.   219  volts.     Calculated,  221  volts  (Chap.  IV.,  Prob.   i). 
Error  0.1%  of  line  voltage.] 

3.  Find  the  size  of  copper  required  for  a  drop  of  6%  or  less  in  the 
following  case: 

Length  of  line 5000  feet. 

Spacing 18  inches. 

Load  (at  receiver  end),  75  Kw.  (79  K.V.A.), 
95%  P.F.,  2000  volts,  single  phase,  60  cycles. 

[Ans.   No.  4  copper.] 

4.  Assuming  No.  4  copper  wire  of  1.312  ohms  per  mile,  find  the 
per  cent  line  drop  and  the  supply  voltage.     (See  Elec.  Journal,  Apr., 
1907,  p.  231.) 

[Ans.    543%,  2110  volts,  calc.  5-43%,  2109  volts  (Chap.  IV., 
Prob.  3)-] 

5.  Find  the  required  size  of  copper  for  9%  regulation  in  the  fol- 
lowing case : 

Length  of  line 25  miles. 

Spacing 3  feet. 

Load  (at  receiver  end),  2500  K.V.A., 

20,000  volts,  60%  P.F.,  three  phase,  25  cycles. 

[Ans.   No.  o  copper.] 

6.  Assuming  No.  o  copper  wire  of  0.520  ohm  per  mile,  find  the 
per  cent  regulation.     (See  Elec.  Journal,  Apr.,  1907,  p.  231.) 

[Ans.  8.42%,  calc.  8.51%  (Chap.  IV,  Prob.  5).    Error  0.09% 
of  line  voltage.] 


REGULATION   CHART  15 

7.  Find  the  per  cent  regulation  and  the  voltage  drop  of  the  fol- 
lowing line: 

Length  of  line 75  miles. 

Spacing,  8  feet,  regular  flat  spacing. 

Conductor No.  oo  aluminum  cable. 

Load  (at  receiver  end),  10,000  K.V.A., 

88,000  volts,  85%  P.F.,  three  phase,  25  cycles. 
[Ans.   7-36%  Reg'n.,  7.15%  line  drop,  calc.  7-35%  Reg'n.,  7.13% 

line  drop  (Chap.  V,  Prob.  3).] 

8.  Find  the  K.V.A.  which  can  be  delivered  at  the  end  of  the  fol- 
lowing line,  with  10%  regulation,  at  75%  and  at  90%  P.F.: 

Length  of  line 100  miles. 

Spacing 10  feet. 

Conductor No.  oooo  aluminum  cable. 

Receiver  voltage 110,000  volts,  three  phase, 

60  cycles. 
[Ans.  14,300  K.V.A.,  16,300  K.V.A.] 


CHAPTER  IV. 
FORMULAS  FOR  SHORT  LINES. 

THE  effect  of  capacity  is  inappreciable  with  short  lines 
as  it  amounts  to  only  y1^  of  i%  for  a  line  about  20  miles 
long.  Thus  distribution  lines  and  many  short  transmission 
lines  can  be  quite  accurately  calculated  without  considering ' 
the  line  capacity  at  all.  The  formulas  in  Tables  I  and  II, 
pp.  1 8  and  20,  enable  one  to  solve  many  problems  con- 
nected with  such  lines. 

The  formulas  are  divided  into  two  groups,  those  in 
Table  I  being  used  when  all  the  particulars  describing  the 
load,  such  as  K.V.A.,  voltage  and  power  factor,  are  speci- 
fied at  the  receiver  end.  Table  II  is  used  when  these  par- 
ticulars are  specified  at  the  supply  end. 

One  first  finds  the  quantities  P  and  <2'or  Ps  and  Qs.  These 
are  the  values  of  in-phase  current,  and  reactive  or  quadra- 
ture current,  at  the  point  where  the  conditions  are  speci- 
fied. It  is  to  be  noted  that  only  values  of  current 
expressed  in  total  amperes  are  to  be  used  in  connection 
with  the  formulas  in  this  book.  The  number  of  amperes 
per  wire  is  never  used  in  the  calculations  (except  with 
single  phase  lines),  but  if  it  is  desired  to  be  known,  it 
may  be  determined  from  the  formulas  at  the  bottom  of 
the  tables. 

The  next  step  is  to  find  the  quantities  A  and  B,  or  F  and 
G.  One  is  then  ready  to  find  the  value  of  any  of  the  ten 
quantities,  whose  formulas  are  given  in  the  tables.  It 

16 


FORMULAS  FOR  SHORT  LINES  17 

should  be  remembered  that  each  of  these  ten  quantities 
may  be  determined  independently  of  all  the  others.  Thus 
it  is  not  necessary  to  work  out  the  first  six  equations  in 
order  to  obtain  the  value  of  the  seventh,  since  the  seventh, 
like  any  of  the  others,  may  be  calculated  directly. 


18  TRANSMISSION  LINE  FORMULAS 

TABLE  I.  —  FORMULAS  FOR  SHORT  LINES. 
CONDITIONS  GIVEN  AT  RECEIVER  END. 

These  formulas  are  exact  when  the  line  is  short.     When  the  line  is  20 
miles  long,  they  are  correct  within  approximately  T^  of  i%  of  line  voltage. 
Conditions  given: 
K.V.A.  =  K.V.A.  at  receiver  end. 

E  =  Full  load  voltage  at  receiver  end. 
cos  6  =  Power  factor  at  receiver  end. 
K.W.  =  K.V.A.  cos  6. 

r  =  Resistance  of  conductor  per  mile.  (From  Tables  VII-VIII.) 
x  =  Reactance  of  conductor  per  mile.  (From  Tables  IX-XII.) 
I  =  Length  of  line  in  miles. 

1000  K.V.A.  cos  6 
Then     P=  —       —  •=  —     —  =  In-phase    current    at    receiver    end    (m 

total  amps.). 

_.       looo  K.V.A.  sin  6      _  f     .. 

Q  =  —     —  -  =  Reactive    current    at    receiver    end    (in 

total  amps.)  when  current  is  lagging 

1000  K.V.A.  sin  6    ,  . 

=  --  =  -  when  current  is  leading. 
hi 

Find  the  following  quantities: 

Three  phase  or  two  phase.  Single  phase. 

A=E  +  Prl  +  Qxl.  A  =  E  +  2  Prl  +  2  Qxl. 

B  =  Pxl-Qrl.  .  B  =  2Pxl-2Qrl. 

Formulas  (capacity  neglected)  : 

(1)  Voltage  at  supply  end  =  A  -\  --  -.  • 

2  A. 

Bz 

(2)  Regulation  of  line  =  A  H  --  j  —  E.    (Same  as  line  drop.) 

2  A. 


(3)  Per  cent  regulation  of  line  =  -  -  -  -  '-  per  cent. 

(Same  as  per  cent  line  drop.) 


(4)  K.V.A.  at  supply  end  = rr^-  X  K.V.A. 

£L 

(5)  K.W.  at  supply  end  =  —  (AP  -  BQ). 

IOOO 

(6)  Power  factor  at  supply  end  =  — (Ap  ~  ^Q)  E 


(in  decimals). 


FORMULAS   FOR  SHORT  LINES 
AP-B 


(7)  In-phase  current  at  supply  end  = 


(8)  Reactive  or  quadrature  current  at  supply  end 


in  total  amperes.* 

When  this  quantity  is  positive,  the  current  is  lagging. 
When  this  quantity  is  negative,  the  current  is  leading. 

(9)  K.W.  loss  in  line  =  -         (AP  -  BQ  -  EP). 


(10)  Per  cent  efficiency  of  line  =    .  „  _  ^  per  cent. 

Total  amps. 
*  Amperes  per  wire,  three  phase  =  --  -j=  -  • 

V3 

Total  amps. 
Amperes  per  wire,  two  phase  =  -  • 


20  TRANSMISSION  LINE  FORMULAS 

TABLE  II.  —  FORMULAS  FOR  SHORT  LINES. 
CONDITIONS  GIVEN  AT  SUPPLY  END. 

These  formulas  are  exact  when  the  line  is  short.     When  the  line  is  20 
miles  long,  they  are  correct  within  approximately  tV  °f  z%  °f  nne  voltage. 

Conditions  given: 

K.V.A.  =  K.V.A.  at  supply  end. 

Es  =  Full  load  voltage  at  supply  end. 
cos  6  =  Power  factor  at  supply  end. 
K.W.  =  K.V.A.  cos  6. 

r  =  Resistance  of  conductor  per  mile.  (From  Tables  VII-VIII.) 
x  =  Reactance  of  conductor  per  mile.  (From  Tables  IX-XII.) 
I  =  Length  of  line  in  miles. 

loco  K.V.A.  cos  9 

Then  Pa  =  -       —  ^  —     —  =  In-phase     current    at    supply    end 

£s 

(in  total  amps.). 

_.        1000  K.V.A.  sin  9      _ 

Q8  =  -  =  -  =  Reactive     current     at    supply    .end 
•c-s 

(in  total  amperes)  when  current  is 
lagging. 

1000  K.V.A.  sin  9     ,  .  .   ,     ,. 

=  --  —  -  when  current  is  leading. 
E8 

Find  the  following  quantities: 

Three  Phase  or  Two  Phase.  Single  Phase. 

F  =  E8-  P8rl  -  Q8xl  F  =  E8-2  P8rl  -  2  &*' 

G  =  Q8rl  -  P8xl  G  =  2  Q8rl  -  2  P8xl. 

Formulas  (capacity  neglected): 

G2 

(1)  Voltage  at  receiver  end  =  F  +  —£• 

(2)  Regulation  of  line  =  E8  —  F  —  —p.     (Same  as  line  drop.) 

2 


(3)  Per  cent  regulation  of  line  =  «        -  per  cent. 


(Same  as  per  cent  line  drop.) 
F+- 


(4)  K.V.A.  at  receiver  end  =  — = X  K.V.A. 

±L8 


FORMULAS  FOR   SHORT  LINES  21 

(5)  K.W.  at  receiver  end  =  ^^  (FP8  —  GQ8). 

i  (FP8-GQg)Es 

(6)  Power  factor  at  receiver  end  =  —  y—      ^  \  x  K.V.A. 


(in  decimals). 
(7)  In-phase  current  at  receiver  end  =  FPs  "  £Qa  in  total  amperes.* 


,      GPs  +  FQS 

(8)  Reactive  or  quadrature  current  at  receiver  end  =  ^r— 

in  total  amperes.* 

When  this  quantity  is  positive,  the  current  is  lagging. 
When  this  quantity  is  negative,  the  current  is  leading. 

(9)  K.W.  loss  in  line  =  ^  (E8PS  -  FPa  +  GQs). 

(10)  Per  cent  efficiency  of  line  =  —        ^p^       —  Per  cent- 

Total  amps. 
*  Amperes  per  wire,  three  phase,  =     — ^= 

_  Total  amps. 
Amperes  per  wire,  two  phase,  —  - 


22  TRANSMISSION  LINE   FORMULAS 

PROBLEM  A. 

„ 

Find  the  regulation  of  the  following  single-phase  line: 
Length  of  line  .......................   15  miles. 

Spacing  .............................     3  feet. 

Conductor  ..........................  No.  o  copper  wire. 

Load  (at  receiver  end),  300  K.V.A.,  11,000  volts,  50%  P.F.,  single 

phase,  60  cycles.     (Same  as  Prob.  D,  Chap.  III.) 
From  the  tables  in  Part  III, 

r  =  0.5331. 
*  =  0.686. 


P  =  looo  X  300  X  0.50  =       6 

11,000 

=  looo  X  300X0.866  = 

11,000 

A  =  11,000  +  2  X  13-64  X  0.5331  X  15 

+  2  X  23.62  X  0.686  Xi5 
=  11,704. 
B  =  2  X  13.64  X  0.686  X  15  -  2  X  23.62  X  0.5331  X  15 

=  "  97' 

Per  cent  regulation  =     I0°    (  11,704  +    97  X  97  --  11,000  ) 
11,000  \  2  X  11,704  / 

=  6.40%. 

•  • 
PROBLEM  B. 

Calculate  the  volts  drop  for  the  following  case,  where  all  the 
conditions  are  specified  at  the  supply,  or  generator,  end  of  the  line: 
Length  of  line  ........................   10  miles. 

Spacing  ..............................     3  feet. 

Conductor  ...........................  No.  2  copper  wire. 

Quantities  measured  at  supply  end:   600  K.V.A.,  6600  volts, 
80%  P.F.,  three  phase,  60  cycles. 

From  the  tables  hi  Part  III, 

r  =  0.8469. 
*  =  0.714. 


FORMULAS  FOR  SHORT  LINES  23 

„  _  looo  X  600  X  0.80  _ 

Pe  ~  6600  ~  72>73' 

n   _  looo  X  600  X  0.60  _ 
V*  —  ~  ~  ~  54-55- 

6600 

F  =  6600  -  72.73  X  8.47  -  54-55  X  7-14 

=  5594-5- 
G  =  54-55  X  8.47  -  72.73  X  7-i4 

=  -  57- 
Voltage  at  receiver  end 


=  5594-5  +  0-3 
=  5595  approximately. 
Drop  in  volts  =  6600  —  5595 

=  1-005  volts. 

Per  cent  drop  =  I0°  X  I0°* 

5595 
=  17.96%  of  receiver  voltage. 

PROBLEMS,  CHAP.  IV. 

(FORMULAS  FOR  SHORT  LINES.) 

1.  Find  the  voltage  drop  in  the  following  case: 
Length  of  line  .........................  3  miles. 

Spacing  ...............................   2  feet. 

Conductor  ............................   No.  o  copper  cable. 

Load  (at  receiver  end),  200  K.V.A.,  2200  volts,  80%  P.F.,  three 

phase,  60  cycles.     (Prob.  2,  Chap.  III.) 

[Ans.    221  volts.] 

2.  Find  (a)  the  P.F.  at  the  supply  end, 

(b)  the  per  cent  efficiency  of  the  line,  for  the  case  in  Prob.  i. 
[Ans.    (a}  78.7%  P-F.     (b)  92.3%  efficiency.] 

3.  Find  the  supply  voltage  in  the  following  case: 

Length  of  line  ........................  ......    5000  feet. 

Spacing  ....................................       18  inches. 

Conductor,  No.  4  copper  wire  of  1.312  ohms  per  mile. 
Load  (at  receiver  end),  75  Kw.,  2000  volts,  95%  P.F.,  single  phase, 
60  cycles.     (Prob.  4,  Chap.  III.) 

[Ans.   2109  volts.] 


24  TRANSMISSION  LINE   FORMULAS 

4.  Find  the  volts  drop  and  the  watts  loss,  in  the  following  line: 
Length  of  line .  . 20  miles. 

Spacing 5  feet. 

Conductor No.  i  aluminum  cable. 

Two  phase,  25  cycles. 

K.V.A.  at  supply  end 10,000. 

Volts  at  supply  end 35,000. 

P.F.  at  supply  end 80%. 

[Ans.    5950  volts,  1770  Kw.j 

5.  Find  the  per  cent  regulation  of  the  following  line: 

Length  of  line 25  miles. 

Spacing 3  feet. 

Conductor,  No.  o  copper  wire  of  0.520  ohm  per  mile. 

Load  (at  receiver  end),  2500  K.V.A.,  20,000  volts,  60%  P.F., 
three  phase,  25  cycles.     (Prob.  6,  Chap  III.) 

{Am.  8.51%.] 


CHAPTER  V. 
K  FORMULAS. 

WHEN  a  transmission  line  is  more  than  20  miles  long, 
the  formulas  for  short  lines  given  in  Chapter  IV  are  no 
longer  accurate,  and  other  formulas  must  be  used,  which 
will  take  into  account  the  capacity  of  the  line.  Such 
formulas,  called  K  formulas,  will  be  found  in  Tables  III 
and  IV,  pp.  28  to  35.  The  same  tables  will  be  found  in 
Part  III  at  the  end  of  the  book. 

The  K  formulas  will  be  found  very  similar  to  those  of 
the  last  chapter,  and  while  they  require  more  arithmet- 
ical work,  they  should  not  be  found  any  more  difficult  to 
understand.  No  more  values  of  line  constants  need  to  be 
looked  up  for  the  K  formulas  than  for  the  " Short  Line" 
formulas.  The  capacity  of  the  line  does  not  enter  into  the 
calculations,  since  its  effect  is  allowed  for  by  means  of  the 
constant  K  which  is  the  same,  at  any  one  frequency,  for 
all  values  of  line  capacity. 

The  formulas  of  this  chapter  assume  that  the  leakage 
current  is  zero;  that  is,  that  no  power  is  lost  from  leakage 
over  the  insulators  or  from  corona.  This  is  a  correct 
assumption  to  make  for  all  voltages  except  the  very  highest 
in  use.  If  it  is  desired  to  make  allowance  for  corona  loss, 
the  formulas  of  Chapter  VI  should  be  used. 

The  accuracy  of  the  K  formulas  is  given  as  approxi- 
mately YO  of  i%  of  line  voltage  for  lines  up  to  100  miles 
long  and  with  regulation  up  to  20%,  and  as  \  of  i%  for 
lines  up  to  200  miles  long,  and  with  the  same  regulation. 

25 


26  TRANSMISSION  LINE  FORMULAS 

These  limits  are  close  enough  for  commercial  work,  so  that 
the  K  formulas  can  be  recommended  for  all  ordinary 
engineering  calculations  of  the  performance  of  long  power 
transmission  lines  under  steady  conditions,  where  the 
corona  loss  is  small.  The  accuracy  of  the  electrical  calcu- 
lations will  be  better  than  the  accuracy  with  which  the 
resistance  and  the  physical  dimensions  of  the  line  are 
generally  known. 

The  K  formulas  are  well  adapted  to  the  solution  of 
long  transmission  lines  which  have  substations  at  inter- 
mediate points  between  the  ends.  In  such  cases  each 
section  of  the  line  between  substations  must  be  calculated 
separately,  beginning  with  the  end  where  conditions  are 
known.  The  first  step  is  to  find  the  voltage,  in-phase 
current  and  quadrature  current  at  the  first  substation. 
The  load  taken  by  the  substation,  expressed  as  in-phase 
current  and  quadrature  current,  must  be  added  to,  or  sub- 
tracted from,  the  above  values  of  current.  When  condi- 
tions are  given  at  the  receiver  end  and  one  is  proceeding 
toward  the  supply  end,  the  substation  load  must  be  added 
to  the  line  load.  When  conditions  are  given  at  the  supply 
end,  the  substation  load  must  be  subtracted  from  the  line 
load,  since  one  is  proceeding  away  from  the  supply.  Hav- 
ing thus  found  complete  conditions  at  one  end  of  the 
second  section  of  the  line,  the  calculation  of  this  section 
may  be  taken  up  in  the  same  way  as  for  the  first  section. 
In  this  manner  the  entire  line  may  be  calculated  and  the 
voltage  and  current  at  the  unknown  end  may  be  deter- 
mined. 

Examples  are  worked  out,  which  will  give  a  clear  idea  of 
the  manner  in  which  the  K  formulas  are  used.  Many 
other  such  examples  have  been  calculated  and  carefully 


K  FORMULAS  27 

compared  with  the  fundamental  formulas.  As  these  ex- 
amples have  covered  the  range  of  practicable  transmission 
lines,  a  sound  basis  is  afforded  for  the  estimate  of  the 
accuracy  of  the  K  formulas  and  for  the  statement  that 
they  are  sufficiently  reliable  for  all  ordinary  engineering 
purposes  in  the  calculation  of  electric  power  transmission 
lines. 


28  TRANSMISSION  LINE  FORMULAS 

TABLE  III.  —  ^  FORMULAS  FOR  TRANSMISSION  LINES. 
CONDITIONS  GIVEN  AT  RECEIVER  END. 

Accurate  within  approximately  TV  of  i%  of  line  voltage  up  to  100  miles, 
and  ^  of  i%  up  to  200  miles,  for  lines  with  regulation  up  to  20%. 

^  =  6(cycles)2      #=2>l6  for  60  cycles.    K  =  0.375  for  25  cycles. 
10,000 

Conditions  given: 

K.V.A.  =  K.V.A.  at  receiver  end. 

E  =  Full  load  voltage  at  receiver  end. 
cos  6  =  Power  factor  at  receiver  end. 
K.W.  =K.V.A.  cos0. 

r  =  Resistance  of  conductor  per  mile.   (From  Tables  VII-VIII.) 
x  =  Reactance  of  conductor  per  mile.     (From  Tables  IX-XII.) 
I  =  Length  of  transmission  line  in  miles. 
Then  P  =  ^°-  —  .    .    .  cos    _  ^    j^gg    current    at    receiver    end 


(in  total  amps.)  . 
Q  =  1000  K.V.A.  sing  =  Reactiye    current    afc    receiver    end 

(in  total  amps.),  when  current  is 
lagging. 

1000  K.V.A.  sin  &      .  L  .   ,     ,. 

=  --  =  -  ,  when  current  is  leading. 


K  FORMULAS 


29 


Find  the  following  quantities: 


Full  Load. 


.       3 


Load. 


i-*(J-Vi. 

\ioooj 


.  —  The  above  are  for  two-  and  three-phase  lines.    For  single-phase 
lines  use  2  r  and  2  x  in  place  of  r  and  #. 


30  TRANSMISSION  LINE  FORMULAS 

TABLE  III.     (Continued.} 

CONDITIONS  GIVEN  AT  RECEIVER  END. 
Formulas: 

Full  Load.  No  Load. 

Voltage  at  receiver  end. 


«:*  (,)&--_£ 


(for  constant  supply  voltage). 
Regulation  at  receiver  end  in  volts,  for  constant  supply  voltage. 


N.B.    The  regulation  at  receiver  end  may  be  expressed  as  a  percentage 
of  E. 

Voltage  at  supply  end. 


(for  constant  receiver  voltage)  . 
Regulation  at  supply  end  in  volts,  for  constant  receiver  voltage. 

f*\  L.    B*  B°* 

(6)     A  H  --  j  —  AQ  --  j-  • 

2  A  2  Ao 

N.B.    The  regulation  at  supply  end  may  be  expressed  as  a  percentage 


K   FORMULAS  31 

Current  at  supply  end  in  total  amperes* 


(7)   VC2  +  D2.  (8)       Co2  +  A2. 

K.  V.A  .  at  supply  end. 


(9)    - 

1000  2A 

.JF".  a/  supply  end. 
(n)  (4C  +  3D).  (12)    ^ 


.          (10)    -^-(^o+^j-  )VCo2+A>2. 


Power  factor  at  supply  end,  in  decimals. 

AC  +  BD  AQC0+B0D0 

(13)    7  -  ™ 
A  + 


In-  phase  current  at  supply  end  in  total  amperes.* 

AC+BD  ,  ,,     A0C0+B0D0 


Reactive  current  at  supply  end  in  total  amperes* 
,    .     BC-AD  ,  Qx     ^o 

(I7)       "' 


When  this  quantity  is  positive,  the  current  is  lagging. 
When  this  quantity  is  negative,  the  current  is  leading. 
K.W.  loss  in  line. 
(19)    —  (AC  +  BD -EP).  (20)   —  (A0C0+  B0D0) 

IOOO  IOOO 

[same  as  No.  12] 
Per  cent  efficiency  of  line. 
,    .       100  EP 


AC+BD 

Total  amps. 
*  Amperes  per  wire,  three-phase,  =  — -p • 

Total  amps. 
Amperes  per  wire,  two  phase,  = 


32  TRANSMISSION  LINE  FORMULAS 

'TABLE  IV.  —  K  FORMULAS  FOR  TRANSMISSION  LINES. 
CONDITIONS  GIVEN  AT  SUPPLY  END. 

Accurate  within  approximately  TV  of  i%  of  line  voltage  up  to  100  miles 
and  \  of  i%  up  to  200  miles,  for  lines  with  regulation  up  to  20%. 

T^      6  (cycles)2      „  ,  ,      ,  „  .. 

K  =  — — .    K  —  2.16  for  60  cycles.    K  —  0.375  for  25  cycles. 

10,000 

Conditions  given: 

K.V.A.  =  K.V.A.  at  supply  end. 

E8  =  Full  load  voltage  at  supply  end. 
cos  9  =  Power  factor  at  supply  end. 
K.W.  =  K.V.A.  cos0. 

r  =  Resistance  of  conductor  per  mile.  (From  Tables  VII-VIII.) 
x  =  Reactance  of  conductor  per  mile.  (From  Tables  IX-XII.) 
/  =  Length  of  transmission  line  in  miles. 

Then  P8  =  — -^~~ =  In-phase  current  at  supply  end  (in 

•c-s 

total  amps.). 

Q8  = '    '       =  Reactive  current  at  supply  end  (in 

•c-s 

total  amps.),  when  current  is  lagging 

1000  K.V.A.  sin  9      .  .   ,     ,. 

= = ,  when  current  is  leading. 


K  FORMULAS  33 

Find  the  following  quantities: 

Full  Load. 


/ 

Vh-i,  _LY. 

3      \ioooy 


=   —  ES  '  -   I  -   I    * 
X     \IOOOj 


.  —  The  above  are  for  two-  and  three-phase  lines.    For  single- 
phase  lines  use  2  r  and  2  x  in  place  of  r  and  #. 


34  TRANSMISSION  LINE   FORMULAS 

TABLE   IV.     (Continued.) 

CONDITIONS  GIVEN  AT  SUPPLY  END. 
Formulas: 

Full  Load.  No  Load. 

Voltage  at  receiver  end. 


(for  constant  supply  voltage). 
Regulation  at  receiver  end  in  volts,  for  constant  supply  voltage. 


N.B.    The  regulation  at  receiver  end  may  be  expressed  as  a  percentage 
of  E. 

Voltage  at  supply  end. 

F+  — 

(4)   E..  (5)  E08= 2^Ea 

F0+^L 

2F0 

(for  constant  receiver  voltage). 
Regulation  at  supply  end,  in  volts,  for  constant  receiver  voltage. 

N.B.    The  regulation  at  supply  end  may  be  expressed  as  a  percentage 
ofEa. 


'  K  FORMULAS  35 

Current  in  total  amperes.*  _ 

(7)   VM*  +  N*  at  receiver  end.          (8)   Vjjf02  +  NQ2  at  supply  end. 
K.V.A. 

(0)   __L_  {p+**.}  VM*  +  N*  at  receiver  end. 
w'    1000  \         2  Fj 

(10)   -   —E8  Vlf02  +  No2  at  supply  end. 

tf.TF. 

(ll)    _J_(FM  +  Gtf)  at  receiver  end.  (12)   -^-  E8M0  at  supply  end. 

7    1000 
Power  factor,  in  decimals. 

FM  +  GN  -  at  receiver  end. 

*  «*'       /  fT2  \ 

f  F  +  ^  ) 


2V2 


In-phase  current  in  total  amperes* 

(IS)   FM  +  ^N  at  receiver  end.  (16)  Jfo  at  supply  end. 

^  +  5 

Reactive  current  in  total  amperes* 
(I7)   GM  ~    -V  at  receiver  end.  (18)  ]V0  at  supply  end. 


When  this  quantity  is  positive,  the  current  is  lagging. 
When  this  quantity  is  negative,  the  current  is  leading. 
K.W.  loss  in  line. 
(IQ)    _I_  (Esp8  -  FM  -  ON).  (20) 


Per  cent  efficiency  of  line. 


(2I)  f      J?p         Per  cent- 

Total  amps. 
*  Amperes  per  wire,  three  phase,  =     — — — 

Total  amps. 
Amperes  per  wire,  two  phase,  =  -   — - — 


36  TRANSMISSION  LINE   FORMULAS 

PROBLEM  A. 

Find  by  the  K  formulas,  the  line  drop  in  the  following  case: 

Length  of  line 100  miles. 

Spacing 8  feet. 

Conductor No.  3  copper  cable. 

Load  (at  receiver  end),  3000  K.V.A.,  66,000  volts,  90%  P.F., 

three  phase,  60  cycles.     (Prob.  A,  Chap.  III.) 
From  the  tables,  r  =  1.078,  x  =  0.840. 

„      1000  X  3000  X  0.90 

P  =  • -f *=  =  40.91. 

66,000 

O  =  looo  X  3000  X  Q.4359  =       o 
66,000 

2  l6 

A  =  66,000  —  66,000  X  — : — 

100 

+40.91  X  107.8(1  --X— ) 
\         3       loo/ 

+  19.81  X84(i-^X  — 
\         6       100 

=  70,580  volts. 

.  1.078  ..  2.16 
B  =  66,000  X  — e-  X  — 
0.840       100 

+  40.91  X  84(1  -\  X^-J  —  19-81  X  107.8  ( i-  -  X  — ) 
\        6       100  /  V       3       I00/ 

=  3150  volts. 


Supply  voltage  =  70,580  +  =  7o,65o. 


Line  drop  =  4650  volts  =  7.05%. 

By  the  fundamental  formulas,  using  the  same  line  constants,  the 
line  drop  is  7.08%  (Prob.  2,  Chap.  VI).    The  discrepancy  is  0.03% 

of  line  voltage. 

PROBLEM  B. 

Find  by  the  K  formulas  the  voltage  at  the  supply  end  of  the 
following  line: 

Total  length  of  line  .........................     300  miles. 

Spacing  ...................................       12  feet. 


K  FORMULAS  37 

Conductor,  266,800  c.m.  aluminum  cable. 

Load  at  receiver  end  of  line,  9000  K.V.A.,  80%  P.F.  (lagging), 

100,000  volts,  three  phase,  60  cycles. 

Load  taken  by  a  substation  at  the  middle  of  the  line,  150  miles 
from  either  end,  2000  K.V.A.  at  the  line  voltage  and  at  70% 
P.F.  (lagging). 
Solution  of  first  section  of  line. 

r  =  0.3410    x  =  0.791     /  =  150, 
P  =  72  Q  =  54. 

From  the  K  formulas, 

A  =  100,000  —  4860  +  3560  +  6360 

=  105,060  volts. 
B  =  2100  -f  8470  —  2670 

=  7900  volts. 
C=  72  -3-50+  1.13  -0.57 

=  69.06  amperes. 
D  =  1.51  -  54  +  2.63  -f  80.59 
=  30.73  amperes. 

A  H — —  =  Ei  =  105,360  volts. 

AC  +  BD 

In-phase  current  =  • — - 

4          I  -^ 


Reactive  current 


=  71.15  amps 
BC  -  AD 


'     2A 

=  —  25.46  amps 
Solution  of  second  section  of  line. 
Conditions  at  middle  of  line : 

Ei  =  105,360. 
In-phase  current  of  substation  load 

looo  X  2000  X  0.70 

= •  =  13.29  amps. 

105,360 

Reactive  current  of  substation  load 

looo  X  2000  X  0.7 141 

— L-a-  =  13.56  amps. 

105,360 


38  TRANSMISSION  LINE  FORMULAS 

s 

PI  =  71.15  -|-  13.29  =  84.44  amps. 
Qi  =  —  25.46  +  13.56  =  —  11.90  amps. 
Then  by  the  K  formulas, 

Ai  =  105,360  —  5120  +  4180  —  1410 
=  103,010  volts. 

BI  =  2200  +  9940  +  590 

=  12,730  volts. 

T?  2 

AI  -\ j-  =  103,800  volts 

2  AI 

=  voltage  at  the  supply  end  of  the  line. 

[By  the  fundamental  formulas,  supply  voltage  =  103,900  volts 
(Prob.  B,  Chap.  VI).] 

PROBLEMS,    CHAP.  V. 

(K  FORMULAS.) 

1.  Find,  by  means  of  the  K  formulas,  the  voltage  drop  from 
the  supply  end  to  the  receiver  end  (the  line  drop)  of  the  following 
line:  « 

Length  of  line 200  miles. 

Spacing 9  feet. 

Conductor No.  ooo  aluminum  cable. 

Load  (at  receiver  end),  4500  K.V.A.,  66,000  volts,  80%  P.F., 
three  phase,  60  cycles. 

Ans.   6650  volts. 
[By  the  fundamental  formulas,  6700  volts.     (Chap.  VI,  Prob.  A).] 

2.  Find  the  regulation  of  the  line  in  Prob.  A,  Chap.  V.     [See 
Prob.  A,  Chap.  III.] 

Ans.   9.37%. 

[By   the    fundamental  formulas   9.40%.      (Chap.  VI,  Prob.  2.) 
Error  0.03%  of  line  voltage.] 

3.  Find  the  per  cent  regulation  and  voltage  drop  of  the  following 
line: 

Length  of  line 75  miles. 

Spacing,  8  feet,  regular  flat  spacing. 

Conductor No.  oo  aluminum  cable. 

Load  (at  receiver  end),  10,000  K.V.A.,  88,000  volts,  85%  P.F., 
three  phase,  25  cycles.     (Prob.  7,  Chap.  III.) 

Ans.  7-35%  reg'n,  7.13%  drop. 


'K  FORMULAS  39 

4.  Find,  by  the  K  formulas,  the  per  cent  voltage  drop,  the  per 
cent  loss,  and  the  power  factor  at  the  supply  end  of  the  following 
line: 

Length  of  line 100  miles. 

Spacing 6  feet. 

Conductor No.  oooo  copper  wire. 

Take  r  =  0.267,  *  —  °-727>  &  =  6.03  X  lo"6. 
Load  (at  receiver  end),   100  amperes  per  wire,   60,000  volts, 
95%  P.P.,  three  phase,  60  cycles.     [Problem  of  Pender  and 
Thomson,  Proc.  A.  /.£.£.,  July,  1911.] 

Ans.    13-09%  drop,  7.61%  loss,  96.58%  P.F. 
[Calc.  by  series,  13.03%  drop,  7.60%  loss,  96.66%  P.F.  (Prob.  4, 
Chap.  VI).] 

5.  Find,  by  the  K  formulas,  the  K.V.A.  and  voltage  at  the  supply 
end,  and  the  efficiency  of  the  following  line: 

Length  of  line  250  Km.  =  155.34  miles. 

Spacing 6  feet. 

Conductor No.  ooo  copper  wire. 

Total  resistance  of  one  conductor  ....  51.5  ohms. 
Total  reactance  of  one  conductor  ....  48 .  o  ohms. 
Load  (at  receiver  end),  15,000  K.V.A.,  86,600  volts,  80%  P.F., 

three  phase,  25  cycles. 

[See  page  91,  "Application  of  Hyperbolic  Functions,"  by  A.  E. 
Kennelly,  University  of  London  Press,  1912.] 

Ans.    15,130  K.V.A.,  97,920  volts,  89.68%. 
[By  series,  15,153  K.V.A.,  97,934  volts,  89.71%.] 

6.  Find  (a)  star  voltage  at  supply  end  at  full  load, 

(6)  star  voltage  at  supply  end  at  no  load, 

(c)  regulation  volts  (star)  at  the  supply  end, 

(d)  amperes  per  wire  at  supply  end  at  full  load, 

(e)  power  factor  at  supply  end  at  full  load, 
(/)  loss  in  line  at  full  load, 

(g)  efficiency  of  the  transmission  line, 

(h)  amperes  per  wire  at  supply  end  at  no  load  (i.e.,  the 

"charging  current"), 
(i)  power  factor  at  supply  end  at  no  load, 

(7)  loss  in  line  at  no  load, 


40  TRANSMISSION  LINE  FORMULAS 

for  the  following  line: 

Length  of  line 300  miles. 

Spacing 10  feet. 

Conductor,  No.  ooo  copper  cable  of  0.330  ohms  per  mile. 
Load    (at  receiver    end),    18,000   K.V.A.,    104,000  line   volts, 

90%  P.F.,  three  phase,  60  cycles. 
[Prob.  A,  page  2,  G.  E.  Review  Supplement,  May,  1910.] 

Answers: 

By  K  Formulas. 

(a)  69,820 

(b)  48,610 

(c)  21,210 

(d)  97.0 

(e)  92.2 
(/)  2530 
(2)  86.5 
(«  91-3 

W      7.1 

0')      940 

7.  Find,  by  the  K  formulas,  the  voltage  at  the  supply  end  of 
the  following  line: 

Total  length  of  line 400  miles. 

Spacing 15  feet. 

Conductor No.  oooo  copper  cable. 

Load  at  receiver  end  of  line,  5000  K.V.A.,  85%  P.F.  (lagging) 

110,000  volts,  three  phase,  60  cycles. 

Load  taken  by  a  substation  at  the  middle  of  the  line,  200  miles 
from  either  end,  2500  K.V.A.,  at  the  line  voltage  and  at  90%  P.F. 

(lagging). 

Ans.   89,720  volts. 

[By  the  convergent  series,  90,190  volts  (Prob.  6,  Chap.  VI).  Error 
0.5%.] 

A  method  of  operation  of  transmission  lines  is  coming 
into  prominence,  by  which  the  voltage  of  the  line  is  kept 
constant  at  all  points,  and  the  inconveniences  due  to  poor 


By  Fundamental 
Formulas. 

Error  in  per  cent 
of  full  voltage 
or  current. 

69,670  volts 

0-3% 

48,950  volts 

0.6% 

20,720  volts 

o.9% 

96  .  59  amps. 

o.5% 

92.35% 

0-2% 

2440  Kw. 

0.6% 

86.90% 

o.5% 

90.97  amps. 

o-5% 

6-47% 

o.7% 

860  Kw. 

0.5% 

K  FORMULAS  4OA 

regulation  are  obviated.  Synchronous  machinery,  con- 
sisting of  either  synchronous  motors  or  generators,  is 
installed  in  the  stations  throughout  the  transmission 
system  in  sufficient  quantity  to  hold  the  voltage  at  a 
constant  value  by  controlling  the  amount  of  leading  or 
lagging  current  supplied  to  the  line.  This  method  will 
probably  come  into  considerable  favor,  for  there  seems  to 
be  practically  no  limit  to  the  extent  of  a  transmission 
system  operated  at  constant  voltage. 

The  following  problem  outlines  the  method  of  calcula- 
tion for  such  cases. 

A  line  400  miles  long  has  a  substation  at  the  middle, 
200  miles  from  either  end.  A  load  of  10,000  Kw.  is 
taken  at  the  receiver  end  of  the  line,  and  8000  Kw.  at 
the  substation.  Find  the  power  factor  which  is  required 
for  these  loads,  in  order  that  the  voltage  at  the  generator, 
substation,  and  receiver  end  may  be  110,000  volts,  the 
following  data  being  given: 

Conductor,  250,000  c.m.  copper  cable,  1 4-foot  spacing, 
3-phase,  60  cycles, 

r  =  0.2284,  x  =  0.813. 

ist  section  of  line,   /  =  200  miles. 
By  the  K  formulas, 

n      1000  X  10.000 

P  =  -  -=90.91; 


Q  is  unknown; 

A  =  110,000  —  9500  +  3910  +  160.3  Q 
=  104,410  +  160.3  Qj 


40B  TRANSMISSION  LINE  FORMULAS 

B  =  2670  +  14,570  -  43-o  Q 
=  17.240  -  43-o  Q- 

Now  the  voltage  at  the  substation  end  of  the  ist  section 
of  the  line  is  110,000;  that  is, 


. 

+-^  =  no,ooo; 

squaring  both  sides,  A2  +  B2  =  121  X  io8. 
This  gives  a  quadratic  equation  in  Q, 

2.754  Q2  +  3198  Q  -  90,150  =  o, 

from  which  ()=  27.53  total  amps.; 


Power  factor  =  =  95.7%, 

94.99 

and,  as  Q  is  positive,  this  is  a  lagging  power  factor. 
The  power  factor  obtained  by  the  hyperbolic  formulas 

is  95-9%- 

Using  the  above  value  of  Q,  we  obtain 

C  =  +82.78, 
D  =  +  90.59. 

In-phase   current   at    substation    end    of    ist    section 

=  +  95-11'- 
Reactive    current    at    substation    end   of   ist    section 

=  ~  77-53- 
2nd  section  of  line,  /  =  200  miles. 


K  FORMULAS  400 


In-phase  current  in  ist  section  =  +95.11, 
In-phase  current  of  substation  load 


1000  X  8000 

-  =  +  72.73 
110,000 


In-phase   current   at   substation   end   of    2nd   section 

=  167.84. 
Reactive   current   at   substation   end   of    2nd   section 


=  Q*-  77-53> 

where  Qx  is  the  unknown  reactive  current  of  the  sub- 
station load. 

By  the  K  formulas,  as  before, 


AI  =  95>3°°  +  J6o.3  Qx, 
Bl  =  32,910  -    43.0  Qx, 

and  voltage  at  generator  end  of  line 


AI-\ l—  =  110,000. 

24l 


From  the  above  we  obtain  as  before  a  quadratic  equa- 
tion in  Qx,  which  gives 

&=+  65.59. 


40D  TRANSMISSION  LINE  FORMULAS 

Power  factor  of  substation  load  =  74.3%,  lagging. 
The  power  factor  obtained  by  the  hyperbolic  formulas 
is  74.6%, 


CHAPTER  VI. 
CONVERGENT  SERIES. 

THE  mathematical  expression  for  finding  the  operating 
characteristics  of  a  transmission  line,  in  which  exact  ac- 
count is  taken  of  all  the  electrical  properties  of  the  line, 
has  been  published  many  times.  It  involves  the  use  of 
hyperbolic  sines  and  cosines,  as  well  as  of  complex  quan- 
tities,* and,  without  some  special  arrangement,  cannot  be 
directly  applied  to  the  calculation  of  a  particular  case. 
For  this  reason,  most  of  the  systems  so  far  published  for 
calculating  transmission  lines  have  used  approximate  for- 
mulas which  have  been  based  on  the  hyperbolic  formulas. 
In  a  few  cases,  an  attempt  has  been  made  to  devise  a 
system  pi  working  which  would  give  the  exact  results  of  the 
fundamental  hyperbolic  formulas,  but  generally  the  labor 
required  in  using  the  systems  is  so  great  as  almost  to  pro- 
hibit obtaining  the  exact  result,  or  else  the  accuracy  of 
the  work  is  seriously  impaired  by  the  necessity  of  inter- 
polating values,  from  tables  of  hyperbolic  functions  which 
have  been  recently  prepared  for  this  purpose  and  are  not 
as  large  and  complete  as  they  should  be  for  good  working. 

The  original  hyperbolic  formulas  can  be  expressed  in 
the  form  of  convergent  series.f  In  this  form  they  do  not 

*  The  hyperbolic  formulas  are  given  in  Chap.  XV. 

f  Prof.  T.  R.  Rosebrugh,  Applied  Science  Magazine,  University  of  To- 
ronto, March,  1909;  Prof.  T.  R.  Rosebrugh,  Proc.  A.  I.E.  E.,  Nov.,  1909, 
p.  1460;  J.  F.  H.  Douglas,  Electrical  World,  April  28,  1910;  Dr.  C.  P. 
Steinmetz,  Electrical  World,  June  23,  1910;  Dr.  C.  P.  Steinmetz,  "Engi- 
neering Mathematics,"  Chap.  V,  1911. 

41 


42  TRANSMISSION  LINE  FORMULAS 

involve  hyperbolic  or  trigonometrical  functions,  and  so  do 
not  require  any  mathematical  tables,  the  only  operations 
being  multiplication  and  addition.  The  series  can  be  car- 
ried to  any  accuracy  desired  by  merely  using  enough  of  the 
terms,  which  diminish  very  rapidly  when  commercial  fre- 
quencies are  involved. 

The  fundamental  formulas  as  expressed  by  convergent 
series  have  been  rearranged,  and  some  new  convergent 
series  have  been  added,  to  make  the  formulas  as  tabulated 
in  this  chapter  directly  applicable  to  the  exact  solution  of 
all  the  problems  treated  by  the  K  formulas.  Exactly 
the  same  final  formulas  in  A,  B,  C,  D,  etc.,  are  used  with 
the  convergent  series  as  with  the  K  formulas. 

Unlike  the  K  formulas,  which  are  expressed  in  .the 
simplest  algebraical  form,  the  convergent  series  involve 
the  use  of  complex  numbers,  that  is,  numbers  containing 
the  well-known  "j"  terms.  No  difficulty  should  be  expe- 
rienced on  this  account,  however,  as  the  rules  for  using 
complex  quantities  are  quite  straightforward,  and  even 
one  who  has  never  worked  with  them  should  be  able  to 
make  use  of  the  formulas  described  in  this  chapter  by 
closely  following  the  instructions. 

Each  of  the  complex  quantities,  (A  +JB),  (P  —JQ), 
Z  =  (r  +  jx)  I*  Y  =  (g  +jb)  lj  etc.,  is  composed  of  two 
parts,  the  first,  a  so-called  "real"  term,  and  the  second,  a 
j  term.  In  adding  complex  numbers,  the  j  terms  must 
be  kept  separate  from  the  others.  Thus 

4  +j  5  added  to  7  +73  =  n  +78. 

In  multiplying  two  complex  quantities,  the  simple  rules 

*  The  notation  Z  =  (r  +  jx)  I,  etc.,  is  used  in  accordance  with  the 
resolution  adopted  by  the  International  Electrotechnical  Commission  in 
Turin,  Sept.,  191 1. 


CONVERGENT  SERIES  43 

of  ordinary  algebra  are  followed,  and  it  must  be  remem- 
bered that 

j-xj-'f 

=  -i, 
and,  therefore, 

-jXJ=+i 


j*  =+J,  etc. 

Thus  (4  +j  5)  X  (7  +y  3)  is  worked  out  as  follows: 

4+75 
7+73 
+  28+jf35 

-  15  +y  12 
+  13+747- 

In  using  the  convergent  series,  E,  P,  and  Q  are  the  same 
as  used  with  the  K  formulas,  E  being  expressed  as  a  real 
number  without  any  j  term.  Z  is  equal  to  (r  -\-jx)  I, 
where  r  and  x  are  taken  from  Tables  VII  to  XII,  Part  III, 
for  resistance  and  reactance  per  mile.  F  is  equal  to 
(g  -\-jb)  I.  The  leakage  conductance,  g,  per  mile,  should 
be  estimated  from  the  most  suitable  test  data  available, 
giving  insulator  leakage  and  corona  loss  under  conditions 
similar  to  those  of  the  line  considered.  The  capacity  sus- 
ceptance,  6,  per  mile,  will  be  found  in  Tables  XIII  to 
XVI,  Part  III. 

After  F  and  Z  have  been  written  down  in  the  form  of 
complex  numbers,  the  product  FZ  should  be  found,  as 
described  above  for  the  multiplication  of  complex  quan- 
tities. From  this  is  obtained 

FZ     FZ  FZ 

-  ,   -—  ,  and   —  , 
24  6 


44  TRANSMISSION  LINE  FORMULAS 

each  expressed  as  a  complex  number  of  a  single  real  term 
and  a  single  j  term.  Multiplying  the  last  two  together 

F2Z2  F2Z2 

gives  -,   from  which    may  be  written 

2-3-4  2.3.4.5 

down.  In  most  cases  no  more  terms  need  to  be  calcu- 
lated, even  for  very  accurate  work,  but  this  is  to  be  de- 
termined while  doing  the  work,  as  one  usually  figures  out 
the  terms  of  these  series  until  they  become  too  small  to  be 

FZ 

considered  when  added  to 

2 

By  addition  of  terms  obtained  above,  the  values  of 

FZ  ,      F2Z2  FZ  F2Z2 

1 h  etc.    and 1 —  +  etc. 

2-3-4  2-3      2-3.4.5 

are  obtained,  each  as  a  complex  number  of  two  terms. 

/FZ  \ 

Multiply  E  by  the  value  found  for  ( —  +  etc.)  and  add 

it  to  E.    Multiply  (P  -  jQ)  by  Z,  or  (r  +jx)  /,  and  by  the 

/Y7  \ 

value  of  I — •  +  etc.)  and   add  it  to   (P  -  jQ)  Z.     The 

above  quantities  are  added  together  to  give  A  -\-jB,  the 
sum  of  all  the  real  parts  being  equal  to  A ,  and  the  sum  of  all 
the  j  terms  being  equal  to  B. 
Similarly,  C  +JD  is  found  by  adding 

(P  -y®,  (P  -JQ)  (^  +  etc.),  EY,  and  EY  (~  +  etc.). 

These  values  of  A,  B,  etc.,  are  inserted  in  equations  i  to 
21  given  with  the  K  formulas,  in  exactly  the  same  way 
as  the  values  of  A,  B,  etc.,  found  according  to  the  second 
page  of  Table  III.  Each  step  of  the  above  procedure  is 
shown  in  the  examples  in  this  chapter. 

The  use  of  Table  VI  is  the  same  as  that  of  Table  V  de- 
scribed above. 


CONVERGENT  SERIES  45 

TABLE  V.  —  CONVERGENT  SERIES  FOR  TRANSMISSION  LINES. 
CONDITIONS  GIVEN  AT  RECEIVER  END. 

The  convergent  series  give  the  results  of  the  fundamental  formulas  as 
accurately  as  desired,  if  a  sufficient  number  of  terms  is  used. 

When  conditions  are  given  at  the  receiver  end,  the  same  as  with  the  K 
formulas,  find  the  quantities: 

Full  Load. 


1     (  P         if)}  7\^ 

2-3-4       2-3.4.5.6              } 
YZ           Y2Z2                    Y3Z3              ^ 

~T  \f        JV)  ^  1  I 

i  •»  n      (  P     iD">  (  T 

'     2  «3     '     2-3.4.5     '     2.3.4-5-6-7' 

-r  JMJ  —  \Jr     JV)  \  i  - 

I                 1                          1                                   ..-     t~  CtC.  1 

\ 

II                      1                               1  •  etc 

1 

1       YZ 

2-3       2-3.4-5       2.3.4-5-6.7 
No  Load. 

V%72                       V373                              \ 
1                                             1     r+r     1 

r 

u  I   ]DV>  —  -c-l  I  |             1 

/        y 

,_l_  inn  —  F.V      T  4- 

2.3.4     2.3.4.5-6         y' 

Z            Y^Z2                   YZZ3                     \ 

f  . 

V         2-3       2-3.4.5       2. 3. 4'5'6. T  I* 

where  Z  =  (r  +y«)  /. 

r  =  resistance  of  conductor  per  mile. 

x  =  reactance  of  conductor  per  mile. 

I  =  length  of  transmission  line  in  miles. 

Y=(g+jb)l. 

g  =  leakage  conductance  of  conductor  per  mile. 

b  =  capacity  susceptance  of  conductor  per  mile. 

Use  A,  B,  C,  D,  etc.,  with  the  equations  in  the  third  and  fourth  pages 
of  Table  HI  to  solve  transmission  line  problems. 

j£2  

Note  i.  —  In  the  formulas,   A  -\ r  is   used   instead   of  VA2  +  B2. 

2  A 

This  approximation  may  be  used  for  very  accurate  work,  as  it  is  correct 
within  approximately  T§^  of  i%  when  the  regulation  is  not  more  than  20%. 
Note  2.  —  The  above  are  for  two-  and  three-phase  lines.  For  single- 
phase  lines  use  2  r  and  2  *  in  place  of  r  and  x,  and  use  |  g  and  |  b  in  place 
of  g  and  b. 


46  TRANSMISSION  LINE   FORMULAS 

TABLE  VI.  —CONVERGENT  SERIES  FOR  TRANSMISSION  LINES. 

CONDITIONS  GIVEN  AT  THE  SUPPLY  END. 

The  convergent  series  give  the  results  of  the  fundamental  formulas  as 
accurately  as  desired,  if  a  sufficient  number  of  terms  is  used. 

When  conditions  are  given  at  the  supply  end,  the  same  as  with  the 
K  formulas,  find  the  quantities: 

Full  Load. 


( 


2-3       2-3.4.5       2.3-4-S 
No  Load. 


+  JG,  =  E.  (i  -  h  YZ  +  £  Y>Z*  -  g  Y'Z*  +  ^  YW  -  etc.), 

,  +jlf,  =  E,Y(I  -$YZ  +  —  Y*2?-^-  YW  +  -,-  Y<Z>  -  etc.), 
\  -"-5  o-'S  2035  / 


where  Z  =  (r  +jx)  I. 

r  =  resistance  of  conductor  per  mile. 

x  =  reactance  of  conductor  per  mile. 

I  =  length  of  transmission  line  in  miles. 

Y=(g+jb)l. 

g  =  leakage  conductance  of  conductor  per  mrle. 

b  =  capacity  susceptance  of  conductor  per  mile. 

Use  F,  G,  M,  N,  etc.,  with  the  equations  in  the  third  and  fourth  pages 
of  Table  IV  to  solve  transmission  line  problems. 

Note  i.  —  In  the  formulas,  F  H  --  -  is  used  instead  of  Vp*  +  G2.    This 

2  F 

approximation  may  be  used  for  very  accurate  work,  as  it  is  correct  within 
approximately  yf  ^  of  i%  when  the  regulation  is  not  more  than  20%. 

Note  2.  —  The  above  are  for  two-  and  three-phase  lines.  For  single- 
phase  lines  use  2  r  and  2  x  in  place  of  r  and  x,  and  use  \  g  and  \  b  in  place  of 
g  and  b. 


CONVERGENT  SERIES  47 

PROBLEM  A. 

Find  the  line  drop,  by  means  of  the  convergent  series,  for  the 
following  line: 

Length  of  line 200  miles. 

Spacing 9  feet- 

Conductor No.  ooo  aluminum  cable. 

Load  (at  receiver  end),  4500  K.V.A.,  66,000  volts,  80%  P.F., 

three  phase,  60  cycles. 
From  the  tables, 

r  =  0.5412,    *  =  0.784,     6=549Xio-6. 
Then  p  =  looo  X  4500  X  0.8  ^^^ 

06,000 
looo  X  45QO  X  0.6  = 

66,000 
Z  =  108.24+.;  156.8. 

Y  = -j-j  0.001098. 

YZ  =  -0.17216  +j  0.11885. 

YZ 

=  —0.08608  +j  0.05942. 

YZ 

—  =  -  0.04304 +y  0.02971. 

4 

YZ 

•  =  —  0.02869 +.70.01981. 

6       

—  0.00059  —.70.00085. 
+  0.00124  —  j  0.00085. 
F2Z2 


=  +  0.00065  —j  0.00170. 
2-3.4 

YZ 

—  =  —  0.08608  +j  0.05942. 

IY7  \ 

h  etc.    =  -  0.08543  +^'0.05772. 

\    2  / 

F2Z2 
=  +  0.00013  —j  0.00034. 

2.3.4-5 
YZ 

—  =  —  0.02869  +y  0.01981. 

2  '3         

( +etc.)=  —0.02856+70.01947. 

\2  •  3  / 


48  TRANSMISSION  LINE  FORMULAS 

Now  E  =  66,000. 

(YZ  \ 

--  h  etc.  J  =  —0.08543  +.;'  0.05772. 

~  +  etc.)  =  -5640  +.7-3810. 


P-JQ=    54-55  -3   40.91- 

708.24  +y  156.8. 


+  5900  -j  4430. 
+  6420+^8550, 

(P  -  JQ)  Z=  +  12320  +  j  4120. 

YZ  \ 

etc.  ]  =  —  0.02856  +  7*0.01947. 
* 


\2«3 


—  80+^*240. 
-350  -./i  20. 


(V7  \ 

—  +  etc. )  =  -  430  +j  120. 
2*3  / 

E  =  66,000. 

E( h  etc.J  =  —  5640+^3810. 

(p-jQ)Z=    12320+^4120. 
(P  -  JQ)  Z>(^-  +  etc.)  =  -    430  +j  120. 


A+jB=  78,320  -  6070+^*8050 
=  72,250+^8050. 

Bz 

A  H =  72,700  volts. 

2  ^1 

Line  drop  =  6700  volts. 

PROBLEM  B. 

Find,  by  the  convergent  series,  the  voltage  at  the  supply  end  of 
the  following  line : 

Total  length  of  line 300  miles. 

Spacing 12  feet. 

Conductor,  266,800  c.m.  aluminum  cable. 

Load  at  receiver  end  of  line,  9000  K.V.A.,   80%   P.F.  (lag- 
ging), 100,000  volts,  three  phase,  60  cycles. 


CONVERGENT   SERIES  49 

Load  taken  by  a  substation  at  the  middle  of  the  line,  1  50  miles 
from  either  end,  2000  K.V.A.,  at  the  line  voltage  and  at  70% 
P.F.  (lagging).     (Prob.  B,  Chap.  V.) 
Solution  of  first  section  of  line: 

r  =  0.3410,    x  =  0.791,     b  =  5.44  X  io~6,    /  =  150. 
Z=  51.15-!-.;  118.65. 
y  =  +j  0.000816. 
YZ  =  —  0.09682  +,70.04174. 

(—  +  -^-  +  etc.  )  =  -  0.04809  +j  0.02053. 

\    2  2  «3  '4  I 

F£__,  --  F2Z2      -I-  etc.)  =  -  0.01607  +j  0.00689. 
2-3       2-3.4-5  / 

E  =  +  100,000. 

E  ~  +  etc    =  -  4810  +j  2050. 


(P-jQ)Z=         10,090  +./5780. 


(P  -  jQ)  Z  (—  +  etc. 


-    200  -    20. 

A+jB=     105,080  +77810. 

Bz 

A  +  —  =  £1  =  105,370  volts. 

In  a  similar  manner,  it  is  found  that 

C+jD  =  69.08  +  j  30.36  amps. 

In-phase  current,  -  =  71.15  amps. 

A+JL; 

2  A 

Reactive  current,  -  -—  —  =  —  25.15  amps. 

A+- 
2  A 

Solution  of  second  section  of  line: 
Conditions  at  middle  of  line, 

Ei  =  105,370  volts. 
In-phase  current  of  substation  load 

1000  X  2000  X  0.70 
=  -  -  L-  =  13.29  amps. 

105,370 

Reactive  current  of  substation  load 

1000  X  2000  X  0.7141 
_  -  i  *  .  =  I3>55  amps< 

105,370 


50  TRANSMISSION  LINE  FORMULAS 

Current  of  substation  load  =  13.29—  j  13. 55. 
Current  of  first  section        =  71.15  -\-j  25.15. 

•Pi  —  JQi =  84.44  +j  1 1. 60. 
£1  =  4- 105,370. 

(YZ  \ 

—  4-  etc.  J  -  5070  4-  J  2, 160. 

(•Pi  ~~  JQi)  Z  =  2>94°  4- J  1 0,6 10. 

(Pi  —JQi)  Z  i-  '-  4-  etc.]  =  -  120  -y  150. 

»•*  *  3  / 

•^i  -}-jBi  =       103,120  4~J  12,620  volts, 

yli  -!> — j-  =  103,900  volts 

=  voltage  at  the  supply  end  of  the  line. 

PROBLEMS,  CHAP.  VI. 

(CONVERGENT  SERIES.) 

1.  Find,  by  the  convergent  series,  the  voltage  drop  of  the  follow- 
ing line: 

Length  of  line 80  miles. 

Spacing 10  feet. 

Conductor No.  oo  aluminum  cable. 

Load   (at   receiver  end),   15,000  K.V.A.,    100,000  volts,    95% 
P.F.,  two  phase,  25  cycles.     (Prob.  C,  Chap.  III.) 

Ans.   8810  volts. 

2.  Find,  by  the  convergent  series,  the  per  cent  line  drop  and  the 
per  cent  regulation  of  the  following  line: 

Length  of  line 100  miles. 

Spacing 8  feet. 

Conductor No.  3  copper  cable. 

Load  (at  receiver  end),  3000  K.V.A.,  66,000  volts,  90%  P.F., 
three  phase,  60  cycles.  (See'Prob.  A.,  Chap.  Ill,  and  Prob.  A, 
Chap.  V.) 

Ans.   7.08%  drop,  9.40%  reg'n. 

3.  Find,  by  the  convergent  series,  the  K.V.A.  and  voltage  at  the 
supply  end,  and  the  efficiency  of  the  following  line: 


CONVERGENT   SERIES  51 

Length  of  line 250  Km.  =  155.34  miles. 

Spacing 6  feet. 

Conductor No.  ooo  copper  wire. 

Total  resistance  of  one  conductor,  51.5  ohms. 

Total  reactance  of  one  conductor,  48.0  ohms. 

Total  susceptance  of  one  conductor,  3.724  X  icf4  mhos. 

Load    (at   receiver    end),    15,000   K.V.A.,    86,600  volts,    80% 

P.F.,  three  phase,  25  cycles.     (Prob.  5,  Chap.  V.) 
Ans.    15,153    K.V.A.,    97,934   volts,   line;     56,542   volts,    star; 

89.71%- 

4.  Find,  by  the  convergent  series,  the  per  cent  voltage  drop,  the 
per  cent  loss,  and  the  power  factor  at  the  supply  end  of  the  following 
line: 

Length  of  line 100  miles. 

Spacing 6  feet. 

Conductor No.  oooo  copper  wire. 

Take  r  —  0.267,  x  —  °-727>  ^  =  6.03  X  io~6. 
Load  (at  receiver  end),  100  amperes  per  wire,  60,000  volts, 
95%  P.F.,  three  phase,  60  cycles.     [Prob.  4,  Chap.  V.] 

Ans.    13.03%  drop,  7.60%  loss,  96.66%  P.F. 

5.  Find,  by  the  convergent  series, 

(a)  star  voltage  at  supply  end  at  full  load, 

(b)  star  voltage  at  supply  end  at  no  load, 

(c~)   regulation  volts  (star),  at  the  supply  end,        »  . 

(d)  amperes  per  wire  at  supply  end  at  full  load, 

(e)  power  factor  at  supply  end  at  full  load, 
(/)   loss  in  line  at  full  load, 

(g)  efficiency  of  the  transmission  line, 

(ti)  amperes  per  wire  at  supply  end  at  no  load  (i.e.,  the 
"charging  current"), 

({)   power  factor  at  supply  end  at  no  load, 

(/)   loss  in  line  at  no  load,  for  the  following  line: 

Length  of  line 300  miles. 

Spacing 10  feet. 

Conductor,  No.  ooo  copper  cable  of  0.330  ohm  per  mile. 
Load  (at  receiver  end),  18,000  K.V.A.,  104,000  volts,  90%  P.F., 
three  phase,  60  cycles.     (Prob.  6,  Chap.  V.) 


52  TRANSMISSION  LINE   FORMULAS 

Ans.  (a)  69,670  volts,  (b)  48,950  volts,  (c)  20,720  volts,  (d) 
96.59  amps.,  (g)  92.35%,  (/)  2440  Kw.,  (g)  86.90%,  (A) 
90.97  amps.,  (i)  6.47%,  (;)  860  Kw. 

6.  Find,  by  the  convergent  series,  the  voltage  at  the  supply  end 
of  the  following  line: 

Total  length  of  line 400  miles. 

Spacing 15  feet. 

Conductor No.  oooo  copper  cable. 

Load  at  receiver  end  of  line,  5000  K.V.A.,  85%  P.P.  (lagging), 

110,000  volts,  three  phase,  60  cycles. 

Load  taken  by  a  substation  at  the  middle  of  the  line,  200  miles 
from  either  end,  2500  K.V.A.  at  the  line  voltage  and  at  90%  P.F. 
(lagging).  (Prob.  7,  Chap.  V.) 

Ans.  90,190  volts. 


PART   II. 
THEORY. 


CHAPTER  VII. 

CONDUCTORS. 

THREE  main  classes  of  conductors  are  used  for  overhead 
lines  for  the  transmission  of  electric  power;  namely,  copper 
wires,  copper  cables  and  aluminum  cables.  The  cables  used 
are  generally  strands  of  seven  wires;  that  is,  they  consist 
of  a  central  straight  wire  with  six 
wires  wound  spirally  around  it,  as 
indicated  by  the  cross  section  in 
Fig.  9. 

From  this  figure  it  is  seen  that  .. 
the  maximum  diameter  of  a  y-wire 
strand  is  equal  to  3  times  the  di- 
ameter of  one  of  the  wires. 

The  outside  wires  do  not  follow 
a  straight  path  parallel  to  the 
central  wire  and  the  axis  of  the  cable,  but  lie  in  a  spiral 
around  it,  as  mentioned  above.  As  there  is  always  a 
slight  insulating  film  of  oxide  on  any  wire,  the  current 
flowing  in  the  cable  tends  to  stay  in  the  individual  wires, 
and  so  follows  the  longer  path.  Thus,  the  resistance  of  a 
cable  is  greater  than  that  of  a  solid  wire  of  the  same  area 
of  cross  section.  The  amount  of  the  difference  depends  on 
the  number  of  wires  in  the  cable  and  the  pitch  of  the 

53 


2s 

Fig.  9.    7-Wire  Strand. 


54  TRANSMISSION  LINE  FORMULAS 

spiralling,  but  an  average  value  of  i%  is  assumed  in  mak- 
ing up  the  tables  in  Part  III.  The  cross  section  of  the 
cable  is  assumed  to  be  equal  to  the  sum  of  the  cross  sections 
of  the  individual  wires.  The  weight  per  unit  length  of  the 
cable  calculated  from  this  cross  section  must  be  increased 
by  the  same  percentage  as  the  above  increase  in  resistance, 
due  to  the  extra  length  of  the  outside  wires.  Since  the 
cross  section  in  Fig.  9  does  not  cut  the  outside  wires  ex- 
actly at  right  angles,  their  sections  as  shown  in  the  figure 

are  really  ellipses,  and  the  di- 
ameter of  the  cable  is  slightly 
greater  than  6  pi.  However, 
this  difference  is  small  and  has 
been  neglected  in  the  figures 
for  diameter  of  cable  tabulated 
in  Part  III. 

The  number  of  wires  in  a 
strand  varies  in   practice  ac- 

j<_ 2$ J       cording  to  the  degree  of  flexi- 
bility and  mechanical  strength 

Fig.  10.    19- Wire  Strand.  J 

desired  by  the  user.  The  num- 
ber of  wires  per  strand  in  the  tables  represents  average 
practice  for  overhead  lines.  The  larger  cables  often  have 
19  or  even  37  wires. 

The  section  of  a  i9-wire  strand  is  shown  in  Fig.  10,  and 
it  is  seen  that  the  maximum  diameter  is  5  times  the  diam- 
eter of  one  of  the  individual  wires.  The  same  increase  of 
i%  in  resistance  is  allowed  as  with  a  7-wire  strand. 

There  is  only  a  very  slight  difference  in  the  reactance 
and  capacity  of  a  7-wire  and  a  i9-wire  strand  of  the  same 
sectional  area,  so  that  values  listed  for  7  wires  may  be 
used  for  19  wires,  and  vice  versa,  without  very  much  error. 


CONDUCTORS  55 

The  resistances  for  direct  current  tabulated  in  Part  III 
have  been  calculated  in  accordance  with  the  recommenda- 
tions of  the  Bureau  of  Standards  for  the  preparation  of  wire 
tables.*  The  Standardization  Rules  of  the*  American  In- 
stitute of  Electrical  Engineers  are  in  agreement  with  these 
recommendations.  According  to  the  Bureau  of  Standards 
and  the  A.  I.  E.  E.,  the  "Annealed  Copper  Standard," 
which  is  of  100%  conductivity,  is  represented  by  a  resis- 
tivity of  0.153022  ohm  per  meter-gram  at  20°  C.  This  is 
equivalent  to  1.72128  micro-ohms  per  centimeter  cube  at 
20°  C.,  assuming  a  density  of  8.89.  This  is  the  same  as 
the  resistivity  of  Matthiessen's  Standard  at  20°  C.,  for- 
merly used  by  the  A.  I.  E.  E.  The  conductivity  of  hard 
drawn  copper  recommended  for  wire  tables  by  the  Bureau 
of  Standards  is  97.3%,  this  value  representing  an  average 
for  good  commercial  copper.  The  average  conductivity 
given  by  the  Bureau  of  Standards  for  hard  drawn  alu- 
minum on  the  centimeter  cube  basis,  assuming  a  density  of 
2.699,  is  60.86%.  The  above  values  have  been  used  in 
preparing  the  tables  in  Part  III,  i%  being  added  to  the 
resistance  for  the  effect  of  spiralling,  as  already  noted. 

If  it  is  desired  to  calculate  the  resistance  of  copper  con- 
ductors for  other  temperatures  than  20°  C.,  the  tem- 
perature coefficient,  a2o,  for  hard  drawn  copper  of  97.3% 
conductivity  should  be  used  in  connection  with  the  formula 

Rt  =  RW  { i  +  «2o  (t  -  20) } 

where  /  is  the  temperature  in  degrees  Centigrade  lor  which 
the  resistance  Rt  is  desired  and  where 

«2o  =  0.00383. 

*  Bulletin  of  the  Bureau  of  Standards,  Vol.  VII,  pp.  71-126,  Washington, 
1911;  Proc.  A.  I.  E.  £.,  Dec.,  1910. 


56  TRANSMISSION  LINE  FORMULAS 

For  other  initial  temperatures  and  other  conductivities, 
temperature  coefficients  should  be  used  as  given  in  the 
table  of  temperature  coefficients  in  Part  III,  which  is 
taken  from  Appendix  E  of  the  Standardization  Rules  of  the 
A.  I.  E.  E. 

For  the  temperature  coefficient  of  hard  drawn  aluminum, 
a  value  of 

«2o  =  0.0039, 

which  is  recommended  by  the  Bureau  of  Standards,  may 
be  used. 


CHAPTER  VIII. 

TRANSMISSION  LINE  PROBLEMS. 

WHEN  conditions  are  given  at  the  receiver,  or  load,  end  of 
a  transmission  line,  the  convergent  series  of  Table  V  give 
at  once  the  voltage,  A  +JB,  and  the  current,  C  +JD,  at 
the  other  end  of  the  line.  By  putting  the  load  current 
equal  to  zero,  we  obtain  the  following  expression  for  the 
no-load  voltage  at  the  supply  end: 

CV7          17272  \ 

i+  —  +  JL^L-  +  etc.) 
2  2-3.4  / 

Thus  the  ratio  of  the  voltages  at  the  two  ends  of  the  line  at 
no  load  is 


E  \          22-3-4 

which  is  independent  of  the  voltage  E,  and  depends  only 
on  the  constants  of  the  line. 

The  absolute  value  of  a  complex  quantity  like  the  volt- 
age AQ  -\-jBo,  is  its  total  numerical  value  independent  of  its 
phase  relation.  This  is  the  same,  in  the  case  of  the  voltage 
AQ  +JB0,  as  its  measured  value,  and  is  equal  to 


or       0  +       - 

2  AQ 


to  a  very  close  approximation  when  BQ  is  smaller  than  A0. 
Since  the  two  complex  quantities  making  up  equation  (i) 
are  equal  in  all  respects,  their  absolute  values  are  eaual, 
and  hence 


CV7         V%72  \ 

i  +  —  +  J-jL-  +  etc.)     (2) 
2        2-3-4  / 

57 


58  TRANSMISSION  LINE  FORMULAS 

When  the  line  is  carrying  full  load,  the  measured  value 
of  the  receiver  voltage  is  E,  and  of  the  supply  voltage, 

B2 

A  H  --  j-   If  the  load  be  thrown  off  and  the  supply  volt- 

2  A. 

B2 

age  be  kept  constant  at  A  -\  --  -  >  then  the  receiver  voltage 

2  A. 

will  rise  to  a  value  EQ.    The  line  is  now  at  no  load,  and  the 
ratio  of  the  voltages  at  the  two  ends  is,  by  equation  (2), 


I        YZ        Y2Z2  \ 

=  absolute  value  off  i  +  —  •  H  ----  (-  etc.  I 

V  2         2  •  3  •  4  / 


A+ 
Thus  Eo  = 


(equation  2,  Table  III). 

We  are  now  in  a  position  to  obtain  the  regulation  of  the 
line,  since  by  the  definition  in  the  A.  I.  E.  E.  Standardiza- 
tion Rules, 

En  —  E 

Per  cent  regulation  =  — *-= —  X  100. 

rL 

Thus,  the  regulation  volts  at  the  receiver  end  which  are 
to  be  expressed  as  a  percentage  of  E,  are 


as  in  equation  3,  Table  III. 


TRANSMISSION  LINE   PROBLEMS 


59 


It  is  often  desirable  to  find  the  regulation  of  a  line  at  the 
supply  end,  that  is,  the  per  cent  change  in  supply  voltage 
from  full-load  conditions  to  no-load  conditions,  when  the 
receiver  voltage  is  kept  constant.  If  the  receiver  voltage 
is  E,  we  have  seen  in  the  preceding  paragraph  that  the  full- 
load  supply  voltage  is  equal  to 


2  A 


and  the  no-load  supply  voltage  is 
E0a  =A0  +  - 


The  per  cent  regulation  at  the  supply  end  is 

Es  —  E0a 

E8 


Xioo, 


and  the  regulation  volts  at  the  supply  end  are,  thus 


77 

Ea  — 


2  A 


A 

—  AQ 


2A 


as  in  equation  6,  Table  III. 

In  the  expression  C  +JD  for  current  at  the  supply  end, 
the    quantity    C    denotes    the 
component  of  current  which  is 
in  phase  with  the  voltage  E  at 
the  other  end  of  the  line.     We 
can,    however,    find    the    com- 
ponent of  supply  current  which 
is   in   phase   with    the    supply 
voltage,    by    first    finding    the     k— 
watts  at  the  supply  end. 

Let  the  supply  voltage  be 

E9  =  A  +JB 


Fig.  n. 


60  TRANSMISSION  LINE  FORMULAS 

and  let  its  phase  be  denoted  by  the  angle  0,  Fig.  n,  where 

7? 

tan  0  =  —  > 

r> 

and,  therefore,  sin  0  = 


^4 

and  cos  0  = 


Similarly,  let  the  current  at  the  supply  end  be  C  +  JD, 
at  a  phase  angle  0,  where 


and,  therefore 
and 


VC2  +  D2 


The  watts  at  the  supply  end  are  equal  to  the  current, 
multiplied  by  the  voltage,  multiplied  by  the  power  factor; 
that  is, 

Watts  =  absolute  value  of  Ia  X  absolute  value  of  Es 

X  cos  (0  -  0) 
=  \/C2+  D2X  VA2+  B2X  (cos  0  cos  0+sin  0  sin  0) 


=  VC2  +  £>2  X 


BD 


as  in  equation  n,  Table  III. 

The  quadrature  volt-amperes,  or  reactive  power,    are 
given  by  the  following  equation  : 


TRANSMISSION  LINE   PROBLEMS  6l 

Reactive  power 

=  absolute  value  of  I8X  absolute  value  of  £8Xsin  (0  —  0). 

=  VC2  +  D2  X  VA2  +  B2  (sin  6  cos  0  -  cos 6  sin  0) 

j     £      x     c__      ^      x     z>  .    ) 

=  £C  -  ylZ). 

When  the  expression  BC  —  AD  has  a  positive  value,  the 
current  at  the  supply  end  is  lagging  behind  the  supply 
voltage,  and  when  the  expression  has  a  negative  value,  the 
current  leads  the  voltage  in  phase. 

We  can  now  obtain  the  in-phase  component  of  current, 
which  is  equal  to  watts  divided  by  voltage  (equations  15 
and  1 6),  and  in  the  same  way  the  quadrature  component 
of  current,  which  is  equal  to  reactive  power  divided  by 
voltage  (equations  17  and  18).  The  power  factor  at  the 
supply  end  is  equal  to  watts  divided  by  volt-amperes 
(equations  13  and  14).  Since  the  power  supplied  is  known, 
being  AC  +  BD,  and  the  power  delivered  at  the  receiver  is 
also  known,  being  equal  to  EP,  their  difference  represents 
the  loss  of  power  in  the  line  due  to  resistance  of  the  con- 
ductors, leakage  over  the  insulators  and  corona  loss. 

The  equations  in  F,  G,  M  and  N  are  quite  similar  to  the 
above  equations  in  their  derivation,  and  they  give  the  solu- 
tions of  similar  problems  when  conditions  are  given  at  the 
supply  end  of  the  line. 


CHAPTER  IX. 
REACTANCE  OF  WIRE,  SINGLE-PHASE. 

Effect  of   Flux   in   Air.  —  Let  there  be  an  alternating 
current,  /,  in  the  transmission  line  wire,  A,  indicated  in 

Fig.  12. 

The  magnetic  field  set  up 
by  the  current  at  P,  a  dis- 
tance x  away  from  the  wire, 
will  be  at  right  angles  to 
the  wire.  The  intensity  of 
the  field  will  be  equal  to  the 
force  on  a  unit  magnetic  pole 
at  P  due  to  the  current  in  the  wire.  The  force  due  to  the 
current  in  a  short  length,  dl,  of  the  wire  will  be 

—  cos  6  =  -  cos  6  d8, 
r2  x 

since  dl  cos  0  =  r  dd 

,  x 

and 


cos0 

The  total  force  at  P  due  to  the  current  in  the  wire  A  is 
equal  to 

1C 

C~2I  cosOdd  ,    ,  .  .v 

(where  re  is  a  constant) 

J  <*        ^ 


TT 

I  sin  0"P 


2/ 

X 

62 


REACTANCE  OF   WIRE,  SINGLE-PHASE  63 

where  /  is  measured  in  absolute  electromagnetic  units. 
When  I  is  in  amperes,  the  field  at  distance  x  is 

-  lines  per  sq.  cm.  (i) 

10  x 

r» s-AV ---! 

_A£ 41 $5. 

_   r-r  -J~.  ^^ 


Fig.  13. 

In  Fig.  13  is  shown  the  cross  section  of  a  single-phase 
transmission  line.  The  lines  of  force  in  the  path  of  thick- 
ness dx  surrounding  the  wire  A  are 

21  j 
dx 

IOX 

per  centimeter  of  the  transmission  line.  These  lines  cut 
the  wire  A  and  produce  an  alternating  voltage  in  it  which 
is  90°  out  of  phase  with  the  current  and  is  equal  to 

2/ 

ju  —  dx  X  icr9  volts, 

x 

where  co  =  2  TT  X  number  of  cycles  per  second,  and  where 
/  is  in  amperes. 

The  voltage  drop  between  the  wires  A  and  B,  due  to 
flux  in  the  air  produced  by  the  current  in  A,  is  obtained  by 
integrating  the  above  expression  from  x  =  p  to  x  =  s. 
The  integration  is  not  carried  beyond  x  =  s,  since  flux 
which  cuts  both  A  and  the  return  wire  B  does  not  produce 
any  voltage  between  them.  The  voltage  drop  is  equal  to 

*«       2  7  s 

jw  — dx  X  io~9  =  y«  2  /  loge  -  X  io~9. 


TRANSMISSION  LINE  FORMULAS 


There  will  be  an  equal  drop  due  to  the  flux  produced  by 
the  current  in  the  wire  B,  so  that  the  total  drop  due  to 
flux  in  air  is 

jco4/loge-  X  lo-9 
p 


volts  per  centimeter  of  line, 


2jw  X  741-1  logio-  X 
p 


(2) 


volts  per  ampere  per  mile  of  single-phase  line. 
Effect  of  Flux  in  the  Conductor.  —  Let  i  be  the  current 

per  unit  area  of  section  at  any  point  in  the  wire  shown  in 

Fig.  14.  (For  the  present 
assume  that  i  is  the  same  at 
all  points  of  the  section.) 

The  total  area  of  section  of 
the  wire  is  irp2  and  therefore 
the  total  current  in  the  wire 
is 

/  =  Trip2. 


Fig.  14.    Section  of  Wire. 


The  total  current  inside  the 
circle  of  radius  x  is 

/l  =  irix2. 

This  is  the  only  current  forcing  flux  around  the  circular 
path  of  width  dxt  since  currents  flowing  nearer  the  surface 
of  the  wire  do  not  tend  to  produce  magnetic  lines  in  a 
path  which  does  not  surround  them.  Thus  the  flux  density 
at  the  radius  x  is 

n     I  _  /•>  /w/f/v*'" 


10  X 


IOX 

2irix 

IO 


REACTANCE  OF  WIRE,  SINGLE-PHASE  65 


The  total  flux  in  the  outer  ring  of  the  section  is 

dx  _  TTJ  (p2  —  x2) 
10  10 

This  cuts  the  element  dx  of  the  wire  and  produces  a  voltage 
along  it  equal  to 

juTri  (p2  —  x2)  io~9  volts  per  cm.  (3) 

This  voltage  leads  the  current  by  90°  in  phase  at  all  sec- 
tions. It  is  greatest  at  the  center  and  zero  at  the  surface 
and  so  is  an  unbalanced  voltage;  it  therefore  causes  a  local 
quadrature  current  to  flow  along  the  center  of  the  wire  and 
return  near  the  surface. 

Let  the  local  current  at  the  element  dx  be  i(X)  per  unit 
area  of  section.  Then  the  average  voltage  drop  along  the 
wire  due  to  the  flux  inside  it  and  the  resulting  local  bal- 
ancing current,  is  equal  to 

juILi  =  juiri  (p2  -  x2)  io~9  +  i(x)  r, 

where  L\  is  the  self -inductance  of  the  wire  due  to  the 
above-mentioned  flux  inside  it,  and  where  r  is  the  specific 
resistance  of  the  metal  in  centimeter  units,  that  is,  the  re- 
sistance of  a  centimeter  cube  of  the  metal.  The  current 
i(X)  adjusts  itself  so  that  the  drop  is  the  same  at  all  parts 
of  the  section.  From  the  last  equation,  we  have 

jwrip^Li         juiri  (p2  -  X2)  9  ,  , 

*(*)  =      —j—  —j—      ~  X  10  y,  (4) 

since  /  =  Trip2. 

As  i(X)  is  a  local  current  in  the  wire,  and  does  not  in- 
crease or  decrease  the  main  current  /,  its  sum  when  added 
up  all  over  the  section  must  be  zero,  and  thus 


r.      _ 
2 1^x^(X)  ax  =  o, 
. 


66  TRANSMISSION  LINE  FORMULAS 

that  is,     j  ^^  fVLi*  -  P2x  icr9  +  x3  icT9)  dx  =  o 
r     Jo 

Now  LI  is  a  constant,  independent  of  x,  and  so 
P4Z,!  -  p4  io~9  +  -  io~9  =  o. 

2 

Therefore  Li  =  \  X  icr9  (5) 

The  voltage  drop  between  the  wires  A  and  B  due  to  the 
flux  inside  both  wires  is 

2JuILi  =  2/w/  X  |  X  lo"9  volts  per  cm. 

=  270)  X  80.47  X  icT6 

volts  per  ampere  per  mile.    The  total  reactive  drop  be- 
tween the  wires  is  thus 

2./«  (80.47  +  741.1  logio-)  X  io~6  (6) 

volts  per  ampere  per  mile  of  single  phase  line. 

This  may  be  written  in  the  following  form  which  is  more 
convenient  for  computation  by  means  of  logarithm  tables: 

Reactance  drop  =  2Ju  X  741.1  Iogi0—  —  X  io~5       (7) 

0.779  p 

volts  per  ampere  per  mile  of  single-phase  line. 

The  above  is  the  usual  formula  for  reactance  of  a  single- 
phase  line.  The  proof  is  longer  than  that  generally  given, 
but  it  has  the  advantage  of  giving  a  correct  idea  of 
the  distribution  of  current  and  magnetic  flux  inside  the 
wire.  As  the  irregular  distribution  of  current  produces  the 
"skin  effect"  described  in  the  next  chapter,  and  necessi- 
tates slight  corrections  in  the  above  formula  for  reactance 
and  in  the  resistance,  the  importance  of  calculating  the 
correct  current  distribution  is  evident.  The  above  formula 
is  sufficiently  accurate,  however,  for  calculating  the  tables 
of  reactance  of  wire  in  Part  III. 


CHAPTER  X. 
SKIN  EFFECT. 

IN  the  last  chapter  a  local  quadrature  current  i(X)  was 
assumed,  whose  resistance  drop  balances  up  the  unequal 
voltages  produced  at  the  center  and  near  the  surface  by 
the  flux  inside  the  wire.  This  local  current,  i(X),  when 
added  up  over  all  parts  of  the  section  of  the'  wire,  amounts 
to  zero,  and  so  cannot  produce  magnetic  lines  in  the  air 
outside  the  wire.  But  it  can  produce  lines  inside  the  wire, 
and  the  effect  of  these  will  now  be  calculated. 

The  reactive  drop  in  one  wire  due  to  the  flux  inside  it 
produced  by  the  main  current  i  is 

juirip2Li  =  j(mif^  X  \  X  io~9  volts, 
where  i  is  in  amperes. 

Then  at  a  distance  x  from  the  center  we  have,  from  equa- 
tion (4),  Chap.  IX, 


X     X  TO-'  -  -          p   _  *     x 
2  r 


10" 


This  is  a  lagging  current  at  the  center  and  a  leading 
current  at  the  surface,  and  it  equals  zero  when  integrated 
over  the  entire  section. 

The  current  i(X) ,  integrated  over  the  circle  of  radius  x}  is 


68  TRANSMISSION  LINE  FORMULAS 

The  flux  density  at  the  element  dx,  due  to  the  above  cur- 
rent, is 

2  Iix\         jwirH  _Q  /         9  ~x 

-  ^  =  *  -  X  10  9  (-  p2X  +  Xs). 

iox        10  r 

The  flux  in  the  ring  outside  of  the  circle  of  radius  x,  due 
to  I(x),  is 

0(x)  -  ^  x  10-9  r  (-^4.  *s)  ^ 

ior  Jx 

jW^io-9/     P4    .    p2*2    .  p4      *4\ 
=  -    —  r-  H  ---  1  ----  J 

ior      \      2          2         4       47 

/COTT2^'  IO~9  /  9    9  4\ 

=  <  -  (-  p4  +  2  p2X2  -  ^4). 

40  r 
This  flux  produces  a  voltage  at  the  element  dx,  equal  to 

U>Vi  IO~18  /  4      ,  9     o  4\  /     \ 

-  (-  p4  •  +  2  p2*2  -  ^4).  (i) 

4  P 

A  local  current,  f(2x),  will  flow  in  order  to  keep  the  volt- 
age drop  uniform  over  the  section.  Let  the  average  drop 
due  to  4>(X)  be 


then 

"Vilo~18  (P4  -  2  p 


Integrate  ^(2x)  over  the  entire  surface  and  as  it  is  a  local 
current 


2irxi(2x)dx  =  o 

r     Jo 


2 


SKIN  EFFECT  69 

—18    /   6  6  " 

Therefore,     /Z^P4  =  —      -  ( -  -  -  + 

2  T         \2  2 

,  .     I    COTTp2  I0~18  f    N 

and  £2  =  —  j (3) 

Thus  i(2x)  =  -  C°7r*I°      (2  p4  -  6  p2*2  +  3  x4)    (4) 

This  current  is  in  phase  with  the  main  current  and,  as  it 
is  negative  at  the  center  and  positive  at  the  surface,  it 
produces  a  stronger  resultant  current  near  the  surface  of 
the  wire.  This  is  the  well-known  "skin  effect."  The 
effect  of  the  quadrature  current  i^  is  to  increase  the  re- 
sultant current  both  at  the  center  and  near  the  surface, 
but  its  effect  is  not  as  large  as  that  of  the  in-phase  current 
i(2x)  and  so  the  net  result  is  a  crowding  of  current  toward 
the  surface. 

The  above  process  may  be  continued  indefinitely,  each 
step  adding  a  smaller  correction  than  the  one  before  to 
the  current  at  radius  x  and  to  the  average  drop  in  the  wire. 

Thus  the  expression  for  i($  x) ,  equation  (4) ,  may  be  inte- 
grated over  the  circle  of  radius  #,  and  will  give  the  value 
of  1(2 x)'  This  current  produces  a  flux  density  at  the  radius 
x,  and  by  integrating  this  over  the  outer  ring  of  the  section, 
the  value  of  0@  X)  is  obtained.  The  flux  <j>(2X)  produces  an 
unbalanced  voltage  which  must  be  corrected  by  a  local 
current  i(3X),  so  as  to  give  a  uniform  drop  over  the  section, 
due  to  the  inductance  Z,3.  Equating  the  total  local  current 
to  zero,  as  before,  gives 

i    coVV  io~27 
LZ  = -     — — 

In  the  same  way  it  is  found  that 

.    i    coVp6  io~36 


70  TRANSMISSION  LINE  FORMULAS 

1  C047T48IO-45 


and  ± 

8640         r4 

Let  the  resistance  of  the  wire  per  centimeter  be  R,  where 
R  =  —  -  ohms  per  cm., 

COTTp2  IO~9 

and  let  m  =  -  -  -  • 

r 

-  ^  IO~9 
R 

Then  the  total  drop  in  the  wire  is 
IR 


(  2  Iog6  -  H  ---  j  — 

\  p         2        J  12 


=  IR  +  ;W  io~9  2  Io6  -  H  ---    —  m 


-  ™*  +  -i2-  ^4  ----  ^       (5) 
48  180  8640  / 

volts  per  centimeter. 
The  total  drop  in  phase  with  the  current  is 


-Y  (6) 

/ 


12  180 

The  total  copper  loss  due  to  all  the  currents  in  the  wire  is 
therefore  equal  to 

PR(I+  -Lm2--Lm4  +  ...V 

V         12  180  / 

This  can  be  checked  by  integrating  the  losses  due  to  the 
total  in-phase  and  quadrature  currents  in  all  parts  of  the 
section  of  the  wire,  the  above  result  being  obtained  by 
this  method  also.  Thus,  in  every  respect,  both  as  to  volt- 
age drop  and  watts  loss,  the  resistance  of  the  wire  to  the 
alternating  current  is 


12 


SKIN  EFFECT  71 

Values  of  R'  for  both  25  and  60  cycles  are  tabulated  in 
Part  III.  When  taking  the  resistance  of  a  conductor  from 
the  tables,  R'  should  always  be  used  for  alternating  current, 
and  R  should  be  used  only  when  the  conductor  carries 
direct  current. 

The  total  drop  in  quadrature  with  the  current  is 


=  /co/io"9)  2log€-  +  -(i  --  m2  -\  --  3-m*—  -  •  •)£    (7) 

(  p        2\          24  4320  I) 

The  series  i  -  —  m2  +  -^-w4  -  •  •  • 

24          4320 

is  thus  a  correction  factor  for  the  term  J  or  80.47  in  the 
ordinary  formula  for  reactance.  Its  effect  is  too  small, 
however,  to  make  any  appreciable  change  in  the  tabulated 
values  of  reactance. 

Proof  by  Infinite  Series.  —  The  above  formulas  for  the 
resistance  and  inductance  of  a  wire  carrying  alternating 
current  are  sufficiently  accurate  for  transmission  line  calcu- 
lations with  ordinary  frequencies.  They  may  .also  be  ex- 
tended to  include  more  terms  without  undue  labor.  How- 
ever, as  skin  effect  formulas  are  generally  obtained  and 
expressed  by  means  of  infinite  series  which  can  be  carried 
out  to  any  degree  of  accuracy  for  high-frequency  work,  a 
short  outline  of  the  derivation  of  the  infinite  series  will  be 
given.  It  will  prove  a  check  upon  the  correctness  of  the 
formulas  given  above,  but  it  will  probably  not  give  as 
clear  an  idea  as  they  do  of  the  actual  distribution  of 
current  in  the  wire. 

Let  an  alternating  current,  7,  of  sine  wave  form  and  of 
steady  value,  flow  in  a  round  wire  of  radius  p.  (See  Fig.  14, 


72  TRANSMISSION  LINE  FORMULAS 

Chap.  IX.)  Let  it  take  up  such  a  distribution  that  the 
drop  at  all  parts  of  the  section  of  the  wire,  due  to  re- 
sistance and  to  magnetic  flux,  is  the  same.  Then  if  i'  be 
the  current  density  at  radius  x,  we  may  assume 

*'  =  flo  +  <*i*2  +  «2*4  +  •  •  •  +  anx2n  +:.--••  (8) 
where  OQ,  a\,  .  .  .  anj  etc.,  are  constants,  independent  of  x. 
(As  the  same  value  of  if  would  be  obtained  for  both  +x 
and  —x,  only  even  powers  of  x  need  be  assumed  for  the 
series.) 

The  total  current  in  the  part  of  the  section  inside  a 
circle  of  radius  x  will  be 


/ 

Jo 


=  /   2irxifdx 


i  .     n..i  ,. 

=  2  TT  [  --  1  ---  r  •  •  •  H  ----  r  •  •  •  r         (9; 
\  2         4  2n  / 

The  flux  density  at  the  radius  x  is 


iox      io  \  2  n 


and  the  total  flux  in  the  outer  ring  of  the  section,  outside 
the  circle  of  radius  x,  is 

2  If 


r» 
=  I 

Jx 


, 
dx 


iox 


io  2          3  n 

The  drop  at  radius  x  due  to  the  flux  0'  is 

jot'  X  io~8 
and  the  resistance  drop  due  to  the  current  at  the  same 


SKIN   EFFECT  73 

part  is  i'r.    Thus  the  total  drop  per  centimeter  of  wire, 
which  is  the  same  at  all  parts  of  the  section,  is 
V  =  jwtf  icr8  +  i'r 


2'  3  *  / 

+  r  (oo  +  aix2  +  a*x*  +  -  -  •  +  an#2n  +•••)"  (n) 

The  above  expression  for  V  is  the  same  for  all  values  of  x, 
and  we  may  therefore  equate  each  coefficient  of  x2,  #4,  etc., 
to  zero.  Thus,  putting 

corrp2  ICT9 

— =  m, 

r 

we  have  a\  = 


2p 


np 

...»     ...     etc. 
and 

V  =  fl^+WTo 


Substituting  the  values  of  ai,  02,  etc.,  we  obtain 
V  —  OQT  -\-jwu  lo"9 

(jm)2a0p2  (jm}n~la^  . 

+' 


(12) 


74 


TRANSMISSION  LINE  FORMULAS 


Now  by  putting  x  =  p  in  the  expression  for  /',  equation 
(9),  we  obtain  the  value  of  the  total  current  in  the  wire, 


2          3 
fgi  +  ^ 


n 


Therefore, 
^ 


n(jm}n~l 


Substituting  this  value  of  OQ  in  equation  (12),  and  putting 


the  resistance  per  centimeter  of  the  wire,  we  obtain 

(JmY  i         ,  (JmY  i 

+...+_+... 


r          j.  j.v 


. 

" 


i    6  \J'"'l      i 

(W 


+ 


_  » 


•    V*Oy 


+ 


This  expression  can  evidently  be  carried  to  any  accuracy 
desired.  It  will  give  the  same  results  as  were  previously 
obtained  in  equation  (5),  by  expanding  the  denominator  as 
a  binomial  of  the  form  (i  +  x)~l  and  multiplying  by  the 
numerator.  This  gives 

V  = 


or, 


0 


io 


~9      - 


This  is  the  voltage  drop,  omitting  the  effect  of  the  flux 


SKIN  EFFECT  75 

outside  the  wire,  and  is  the  same  as  the  value  previously 
obtained.     (See  equations  6  and  7.) 


REFERENCES. 


Maxwell,  Elec.  and  Magn.,  Vol.  II,  Para.  689-690. 
Rayleigh,  Phil.  Mag.,  1886,  Vol.  21,  page  381. 
Kelvin,  Math.  Papers,  1889,  Vol.  3,  page  491-       * 
Rosa  and  Grover,  Bulletin  of  Bureau  of  Standards,  Washington,  1911, 
Vol.  8,  No.  i,  pages  173-181. 


CHAPTER  XL 

REACTANCE  OF  CABLE,  SINGLE-PHASE. 

As  stranded  cables  are  very  commonly  used  for  trans- 
mission lines,  it  is  desirable  to  have  a  special  formula  for 
the  reactance  of  cables.  An  outline  will  be  given  of  the 
method  of  obtaining  the  formula  for  a  seven-wire  cable. 
This  formula  was  used  in  preparing  the  reactance  tables 
in  Part  III. 

A  seven-wire  strand  consists  of  a  central  straight  wire, 
with  six  wires  of  the  same  size  laid  around  it  in  a  spiral. 
The  spiralling  of  the  wires  increases  the  resistance  of  the 
cable  by  an  amount  which  is  taken  as  i%.  The  spiralling 
also  increases  the  outside  diameter  of  the  cable  by  about 
TV  of  i%.  (See  Chapter  VII.) 

In  calculating  the  reactance  of  the  cable,  the  first  step  is 
to  plot  the  flux  density  at  various  distances  from  the  center 
of  the  cable.  (See  Fig.  15.)  For  points  entirely  outside 
the  cable,  the  flux  density  obeys  the  law 

27 


where  x  is  the  distance  from  the  center.     The  total  voltage 
due  to  these  lines  which  cut  the  entire  cable  is 

2loge-  X  io~9 
p 

volts  per  ampere  per  centimeter.    When  x  is  less  than  the 
radius  of  the  cable,  the  flux  at  the  distance  x  is 

' 


76 


REACTANCE  OF   CABLE,  SINGLE-PHASE 


77 


I(X)  is  proportional  to  the  area  of  conductor  inside  the 
circle  of  radius  x,  and  this  must  be  measured  from  the 
diagram  of  the  section  of  the  conductor  (Fig.  15)., 


/' 

~ 

x 

^^ 

^ 

•  — 

"-* 

-, 

s' 

'^ 

* 

^^ 

-^ 

/ 

/- 

x^ 

X 

x~ 

'  —  -* 

•^ 

^> 

-^ 

^ 

^ 

^x 

^ 

0       .1       .2       .3       .4       .5       .6       .7       .8       .9        1.0     I.I 
X 

p 

Fig.  15. 

Plot  a  curve  of  flux  density,  /(x),  for  various  values  of  #. 
Let  0(x)  be  the  area  of  the  curve  of  /(x)  between  the  values 
#  and  p,  where  p  is  the  outside  radius  of  the  cable.  Also, 
let  C(X)  be  the  length  of  that  part  of  the  circle  2  TX  which 
lies  in  the  section  of  the  wires  of  the  cable. 

As  shown  in  Chapter  IX,  page  65,  the  voltage  drop 
along  different  parts  of  the  cable  is  not  uniform,  and  must 
be  balanced  by  a  local  quadrature  current.  Thus  we  have 
the  conditions 


io 


~8 


at  any  section, 


78  TRANSMISSION  LINE  FORMULAS 

and  2X'z)£(z)  dx  =  o. 

o 

Substituting  the  value  of  i(X)  from  the  first  equation,  we  have 


dx  ->zCz  dx  io~8  =  o. 
<r 

Now  I  LI  is  a  constant,  and 

2.c(X)dx  =  A, 
(T 

the  area  of  the  section  of  the  cable. 

By  dividing  p  into  a  number  of  equal  parts  and  calling 
each  part  dx,  the  value  of  4>(X)C(X)  dx  for  each  successive 
value  of  x  is  found.  Adding  these  together,  a  close  estimate 
of  the  value  of 


dx 

is  obtained.    This,  divided  by  I  A  gives  the  value 

Li  =  0.633  X  icr9. 

If  the  conductor  were  a  solid  wire  of  radius  p,  with  a 
straight  line  curve  of  /(x),  £1  would  be  J  X  lo"9,  which 
would  become  80.5  X  io~6  when  reduced  to  the  ordinary 
formula  for  reactance  per  mile.  Thus  the  reactance  per 
mile  of  cable  is 

x  =  (80.5  X  ^3  +  74I.!  loglo  -)  io-«  X  2  irf 
\  0.500  p/ 

=  (  102  +  74I.I  logio-j  I0~6  X  2  7T/ 

ohms  per  mile. 

The  same  process  applied  to  a  ig-wire  strand  gives  the 
formula 

x  =  ($9  +  741.1  logio-j  icr6  X2irJ 
ohms  per  mile. 


CHAPTER  XII. 
REACTANCE  OF  TWO-PHASE  AND  THREE-PHASE  LINES. 

Reactance,  Two-phase.  —  The  reactive  drop  in  a  single- 
phase  line,  in  which  round  wire  is  used,  is 

2JI  X  2  7r/f  80.5  -f  741.1  log-)  io~9 

volts  per  mile  of  line.  In  a  two-phase  four-wire  line,  the 
drop  will  be  the  same  as  the  above,  when  /  is  the  current 
in  one  phase. 

AT  T          K.V.A. 

Now  2  /  =  — - —  =  the  total  amperes. 

hi 

Therefore  the  reactive  drop  per  mile  of  line,  in  absolute 
value,  is 

^|^  X  2  T/(8o.s  +  741.1  log-)  io-' 
£  \  p/ 

K.V.A., 

~E~~Xx> 

wnere  x  is  the  tabulated  value  of  reactance. 

Reactance,  Three-phase,  Irregular  Spacing.  —  When  the 
conductors  of  a  three-phase  line  are  spaced  so  that  they  are 
not  exactly  equidistant,  the  voltage  drop  due  to  reactance 
is  not  the  same  in  the  different  phases.  It  is  the  practice 
with  such  lines  tp  interchange,  or  transpose,  the  conductors 
at  intervals  along  the  line,  so  that  the  different  reactive 
voltages  are  applied  to  an  equal  extent  to  all  three  con- 
ductors. Such  a  line,  when  carrying  a  balanced  load  (equal 
currents  in  each  conductor,  at  120°  in  phase  from  each 

79 


8o 


TRANSMISSION  LINE  FORMULAS 


other),  will  have  the  voltages  of  the  three  phases  equal  at 
the  end  of  the  line. 

The  average  reactance  of  an  irregularly  spaced  three- 
phase  line,  in  which  the  conductors  are  transposed  at  regu- 
lar intervals,  may  be  calculated  as  follows: 


Fig.  i 6. 


17. 


Let  Fig.  1 6  represent  the  spacing  of  the  conductors  A,  B, 
and  C  of  a  transmission  line.  Let  the  currents  in  the  con- 
ductors be  represented  by  the  vectors  OP,  OQ,  and  OR 
(Fig.  17).  If  the  power  factor  is  100%,  these  vectors  may 
also  represent  the  star  voltages,  and  the  line  voltages  will 
be  represented  by  the  vectors  PQ,  QR,  and  RP. 

Let  the  current  in  conductor  A  be 


and  let 
and 


I  A  =  OP  =  i  .00  /  amperes, 

IB  =  OQ  =  (—  0.50  +  o.S66j)  I  amperes, 

In  =  OR  =  (  —  0.50  —  0.866 j)  /amperes. 


Let  also 

Voltage  from  neutral  to         A  =  i.oo  V. 

Voltage  from  neutral  to          B  =  (-  0.50  +  0.866.7)  V. 

And  voltage  from  neutral  to  C  =  (  —  0.50  —  0.866.7")  V. 


Then  the  measured,  or  absolute,  value  of  voltage  between 
B  and  C  is  1.732  V,  where  V  is  the  star  voltage  of  the  line. 


REACTANCE  OF  TWO-PHASE  AND  THREE-PHASE  LINES  8  1 
The  reactive  voltage  on  conductor  A  per  mile  is 

1,00/80.5  +  74I.I  log—  \Ij  2  7T/  X  IQ-6 

(due  to  flux  from  7^) 
+  (-0.50  +0.8667)  (741-1  logyJ//27r/X  io~6 

(due  to  flux  from  IB) 
+  (-  0.50  -  0.866  j)  (741.1  logy  JIj  2  TT/  X  io~6 

(due  to  flux  from  Ic)j 

where  p  is  the  radius  of  the  wire. 
The  reactive  voltage  on  conductor  B  per  mile  is 

1.00(741.1  log  —  J7;  2  irf  X  io~6 
(due  to  flux  from  I  A) 


(due  to  flux  from  IB) 
+  (—  0.50  —  0.866  j)  (741.1  log  —  J  Ij  2  irf  X  I 

(due  to  flux  from  /<?). 

The  reactive  voltage  between  A  and  B  is  therefore 
(1.50  -  o. 

+  (0.50  +  0.866  j)  741.1  (log-  -  log-)//  27T/  X  iQ-6. 
\      p  pi 

Suppose  at  the  end  of  one  mile  the  line  is  transposed 
so  that  the  above  two  conductors  occupy  the  positions  B 


82  TRANSMISSION  LINE  FORMULAS 

and  C.    Then  the  reactive  voltage  between  these  conductors 
for  the  next  mile  is 


(1.50  -  0.8667)  (80.5  +  74i.i  log-)  Ij  2  TT/  X  iQ-6 

+  (0.50  +  0.8667)  74LI  (log-  -  log-)  Ij  2  7T/  X  I0~6. 

\     p          p/ 

Let  the  line  be  transposed  again.  Then  for  the  third 
mile  the  reactance  voltage  between  these  same  two  con- 
ductors is 

(1.50  -  0.8667)  (80.5  +  741.1  log  -}  Ij  2  TT/  X  lo-6 

+  (0.50  +  0.8667)  74I.I  (log-  -  log-)  Ij  27T/X  IO-6. 

\      p  P/ 

The  total  reactive  voltage  for  the  three  miles  is 

(1.50-0.8667)  j  80.5X3  +  741.1  (log-  +  log-  +  log-)  I 
(  \      P  p  P/  ) 

X   Ij  2  7T/  X   XT*. 

/Thus  the  average  reactive  voltage  per  mile,  which  is  the 
same  for  all  phases,  is,  in  absolute  value, 

1.732(80.5  +  741.1- log-      -J/27T/X  IO-6. 

\  P    / 

Now  1.732  7  =  the  total  amperes. 

_  K.V.A. 

E 
Therefore  the  drop  in  line  voltage,  in  absolute  value,  is 

(8o.5  +  74i.ilog^27T/Xio-6 

in  volts  per  total  ampere  per  mile  of  transmission  line, 
where 

s  =  ^  dbc. 


REACTANCE  OF  TWO-PHASE  AND  THREE-PHASE  LINES   83 

Thus  the  total  reactive  drop  is  equal  to 

K'VA-  (80.5  +  741.1  log      2  TT/  X  io-6 


E 

K.V.A. 


Xx, 


E 

where  x  is  the  tabulated  value  of  reactance. 

Reactance,  Three-phase,  Regular  Spacing.  —  When  the 
conductors  are  spaced  at  the  corners  of  an  equilateral  tri- 
angle of  side  5,  then  the  expression  for  reactance  is  the 
same  as  the  usual  formula: 


x  =  (80.5  +  741.1  logio-J  27T/ 


X 

Reactance,  Three-phase,  Flat  Spacing.  —  When  the  three 
conductors  lie  in  one  plane  (either 

horizontal  or  vertical),  the  center  j  -  ?  -  j 
one  being  equidistant  from  the  k--—  a—  -4-  .....  a  —  -»! 
other  two,  as  in  Fig.  18,  the  react-  Fig.  18. 

ance  per  mile  is 

i     a  "^i  X  i  X  2 
80.5  +  741.1  log  -  - 
P 

,     1.260 

=  80.5  +  741.  i  log  --  . 
P 

This  is  approximately  4%  higher  in  an  ordinary  case 
than  the  reactance  for  spacing  on  an  equilateral  triangle 
of  side  a. 

The  formulas  of  this  chapter  will  apply  to  cable  as  well 
as  to  wire,  if  the  term  80.5  is  changed  as  per  Chapter  XI. 


CHAPTER  XIII. 


CAPACITY  OF  SINGLE-PHASE  LINE. 

Capacity  of  Two  Round  Wires.  —  The  conductors  of  a 
transmission  line  form  a  condenser,  the  electrostatic  capac- 
ity of  which  can  be  calculated  from  the  dimensions  of  the 
line.  The  simplest  line  to  calculate  is  a  single-phase  line 
consisting  of  two  round  wires,  and  this  case  will  be  in- 
vestigated first. 


Fig.  19. 


Fig.  20. 


Suppose  that  A  and  B  (Fig.  19)  are  two  long  parallel 
wires  of  infinitesimal  section  and  that  they  are  spaced  a 
distance  t  centimeters  apart.  Let  A  carry  a  charge  of  +  q 
electrostatic  units  of  electricity  per  centimeter,  and  let  B 
carry  —  q  units  per  centimeter. 

First,  find  the  force  exerted  on  a  unit  charge  near  the 
wire  A . 

From  the  symmetry  of  the  arrangement  it  is  evident  that 
the  resultant  force  on  a  unit  charge  at  P  (Fig.  20)  will  be  a 
repulsion  away  from  the  wire  at  right  angles  to  it,  since  the 
total  effect  of  the  right  half  of  the  wire  must  be  equal  to 
the  total  effect  of  the  left  half.  The  force  at  right  angles 

84 


CAPACITY  OF   SINGLE-PHASE  LINE  85 

to  the  wire  exerted  by  the  charge  on  the  element  dl  will  be 


=  -^  cos  0  dd, 

r\ 


since  dl  cos  0  =  r  dB 

V, 

and  r  = 


cos  0 
The  total  force  exerted  by  the  wire  will  be 


The  potential  of  the  point  0  (Fig.  19)  midway  between 
the  wires,  will  be  zero,  since  the  effect  of  the  positive  charge 
on  A  will  be  equal  to  the  effect  of  the  negative  charge  on  B. 
The  potential  difference  between  P  and  O  is  the  work  done 
in  moving  a  unit  charge  from  one  point  to  the  other.  The 

2  Q 

force  due  to  the  wire  A  on  a  unit  charge  at  any  point  is  —  *  , 

acting  directly  away  from  A.    Therefore  the  work  done 
in  moving  a  distance  dr  toward  A  is 


and  thus  the  total  work  in  moving  from  P  to  0  against  the 
force  due  to  A  is 

i_ 

r~2Q,                                t 
—-*•  dr  =  20  loge 
0  r                       2ri 

Similarly,  the  work  against  the  force  due  to  B  is  equal  to 

2r2 


86  TRANSMISSION  LINE  FORMULAS 

Therefore  the  potential  difference  between  P  and  0  is 
equal  to  the  total  work,  and  is 

2  q  loge  -  • 


Fig.  21. 

At  P  (Fig.  21)  make  the  angle  APD  equal  to  the  angle 
PBD.  Then  the  triangle  PBD  is  similar  to  the  triangle 
APD,  and  therefore 

PB  ^r* 

AP  "  n 


p 

If  we  draw  a  circle  of  radius  p  about  the  fixed  point  D, 
then  at  any  point  on  this  circle  similar  triangles  are  formed 
by  p,  r\j  and  r^  as  in  Fig.  21,  and  therefore 

r2      DB 

—  =  —  =  constant. 
ri        p 

where  r\  and  r2  are  the  distances  of  the  point  on  the  circle 
from  A  and  B  respectively.     Therefore,  the  potential 

2#log€- 
fi 

will  be  the  same  at  all  points  on  this  circle. 

Now  let  a  solid  cylindrical  conductor  fill  all  the  space 
inside  the  circle  of  radius  p.  All  points  on  its  surface  will 
be  at  the  same  potential.  The  distribution  of  potential 
outside  of  the  cylinder  will  not  be  altered  from  the  previous 
condition  when  all  points  on  the  circle  of  radius  p  were 


CAPACITY  OF  SINGLE-PHASE  LINE  87 

also  at  the  same  potential.    The  potential  of  the  cylinder 
will  be 

,      W  ,       DB 

2qloge-  =  2qloge 

r\  P 

In  the  same  way,  let  the  wire  B  be  replaced  by  a  solid 
cylinder  of  radius  p  and  center  E,  as  in  Fig.  22. 


—  s  — 

...  t  .. 

Fig.   22. 


The  potential  of  this  cylinder,  which  carries  —  q  units 
per  centimeter,  will  be 

AE  ,      DB 


P  P 

since  the  second  cylinder  is  symmetrical  with  the  first. 
Thus  the  potential  difference  between  the  two  cylinders  is 

DB       a 

4?k)ge =  *, 

P          v 

where  C  is  the  capacity  per  centimeter  of  line. 

ThuS      If         \       C=—^DB- 

4log€ 

P 

DB      FB 

Now  -  =  •— , 

p        FA 

since  F  is  a  point  on  the  circle. 

DB          DB-  p 


rpr,  ( 

Therefore 


p        DB  +  p  -  s 

where  5  is  the  interaxial  distance  between  the  two  solid 
cylinders,  and  is  equal  to  DE,  Fig.  22. 


88  TRANSMISSION  LINE  FORMULAS 

Therefore, 

DB2  +  DB  -p  -  DB  -s  =  DB  -p  -  p2 
or  DB2  -  DB  •  s  +  p2  =  o. 

Solving  this  quadratic  equation  in  DB,  we  have 


2 

[The  negative  value  of  the  radical  must  not  be  used, 
since  it  would  give  the  value  of  DA  instead  of  DB.] 
Therefore, 

i 


$4V£-' 

\  2  p        *    4  p 
which  may  be  expressed  as 


4  cosh"1  ( — ) 

\2p/ 

or  it  may  be  expanded  by  the  Binomial  Theorem  to  give 
the  approximate  value 

T 

per  centimeter, 


or,  very  nearly, 

4loge- 
P 

*  It  is  evident  that  the  expression 

i 


C  = 


P 

which  is  sometimes  published,  is  less  accurate  than  the  simpler  expression 


4  log,- 


CAPACITY  OF  SINGLE-PHASE  LINE  89 

Transferring  to  other  units, 

r  =  T  x  *4343  X  2'54°  X  I2  X  528° 

"  4X9XXO" 


x 

2 


loglo(£_£) 
V      s/ 

38.83  x  io-9 

,  (S         p\ 

logio  ---) 
Vp      s/ 

farads  per  mile  of  single-phase  line. 
The  capacity  susceptance  is 

38.83  X 


fr  f  v     N/ 

2  7T/  C   =    2  7T/  X  -   X 


mhos  per  mile  of  single-phase  line.     The  charging  current 
in  this  line  will  be 


amperes, 


where  b  is  the  tabulated  value  of  capacity  susceptance. 

REFERENCE.  —  "A  Treatise  on  the  Theory  of  Alternating  Currents,"  by 
Alexander  Russell,  1904,  Vol.  I,  page  99. 

Capacity  of  Cable.  —  The  formula  for  capacity  of  a  line 
using  stranded  cables  will  be  the  same  as  the  above  formula 
for  solid  wires,  p  being  taken  as  the  maximum  radius  of 
the  cable.  All  the  electrostatic  charge  on  the  cable  does 
not  lie  at  the  maximum  radius  from  the  center,  but  as 
actual  cables  are  generally  slightly  larger  than  the  calcu- 
lated diameters  in  the  tables,  it  will  be  sufficiently  close  to 
take  p  from  the  tables  and  use  it  in  the  regular  formula  for 
capacity. 


90  TRANSMISSION  LINE  FORMULAS 

Effect  of  the  Earth  on  Capacity  of  Line.  —  The  effect  of 
bringing  a  conducting  plane,  such  as  the  earth,  near  to  two 
charged  wires  is  to  change  their  electrostatic  field  and  in- 
crease their  capacity. 

Consider  two  long  parallel  wires,  A  and  Al  (Fig.  23),  of 
infinitesimal  section  and  carrying  +  q  and  —  q  units  of 

electricity  per  centimeter 
respectively.  As  in  Fig. 
19,  the  point  O  midway 

M  __      /        o  N    between     the     two    wires 

will  be  at  zero  potential. 
All  points  having  the  same 

potential  must  have—  equal 

to  a  constant.  It  is  evi- 
dent that  all  points  at  the  same  potential  as  0  lie  in  the 
plane  MON,  perpendicular  to  AA1}  since  for  all  such 
points 


Therefore,  the  wire  A±  may  be  replaced  by  a  solid  con- 
ducting plane  MN,  which  will  be  at  zero  potential.  Thus, 
when  the  conducting  plane  is  the  earth,  its  effect  is  the  same 
as  that  of  a  charged  wire  at  a  depth  below  the  surface 
equal  to  the  height  of  the  original  wire.*  The  assumed 
wire  is  called  an  image  wire,  since  it  occupies  the  same 
position  as  the  image  of  the  real  wire,  considering  the 
surface  of  the  ground  as  a  mirror. 

In  the  case  of  a  single-phase  transmission  line,  image 
wires  A'  and  B'  must  be  assumed  for  both  wires  A  and  B 

*  See  "  Elements  of  Electricity  and  Magnetism,"  by  J.  J.  Thomson,  page 
138. 


CAPACITY  OF  SINGLE-PHASE  LINE 


A' 


----> 

-, — ( 


(Fig.  24)  and  the  capacity  of  the  entire  system  of  four  wires 
is  then  calculated  as  follows: 

Let  h  be  the  distance  of  the 
wires  from  the  ground  and  5 
their  distance  apart  from  center 
to  center.  Let  A  carry  a  charge 
of  +  q  units  per  centimeter;  A',  7 
—  q  units;  B,  —  q  units;  and 
B' ,  +  q  units.  Let  a  unit  charge 
be  carried  from  the  surface  of 
A  to  that  of  B.  Assuming  that  Fig  24 

the  charges  are  concentrated  at 
the  centers  of  the  wires,  the  total  work  done  is  equal  to 


/ 

Jp 


-I 


2  q  (s  -  x) 


dx. 


The  sum  of  the  first  two  integrals  has  been  shown  to  be 
approximately 


The  sum  of  the  last  two  integrals  is  approximately 
4/r>-f  s2 


2*log€i¥T7- 

Therefore,  the  total  work  is  equal  to 
£  ,         ,  zh 

P 


92  TRANSMISSION  LINE  FORMULAS 

Therefore,  C  is  approximately 


!          S     .  2  k 

4loge-  + 

P 


Taking  as  an  average  case, 

h  =  360  inches  (30  feet), 
s  =  120  inches  (10  feet), 
p  =  0.25  inch, 

we  have  -  =  480, 

p 

2  h      . 

and  —  =  6. 

s 


Therefore,       —-  = 
\/4  h2  +  s2 

Now,  Iogio48o  =  2.681, 


while  logio  —  r^  =  0.006. 

6 

Thus  the  capacity  is  changed  by  the  nearness  of  the 
ground  by  less  than  }  of  i%,  even  with  the  comparatively 
wide  spacing  of  10  feet. 

Tests  have  shown  that  the  effect  of  the  ground  in  increas- 
ing the  capacity  is  even  less  than  the  above  amount,  due 
partly  to  the  fact  that  the  ground  is  a  poor  conductor. 
As  the  effect  of  the  ground  is  so  slight,  it  has  been  neglected 
entirely  in  the  calculations  in  this  book. 

REFERENCE.  —  For  an  alternative  proof,  see  "A  Treatise  on  the  Theory 
of  Alternating  Currents,"  by  Alexander  Russell,  1904,  Vol.  I. 

See  also  "  The  Calculation  of  Capacity  Coefficients  for  Parallel  Suspended 
Wires,"  by  Frank  F.  Fowle,  Elec.  World,  Aug.  19,  1911. 


CHAPTER  XIV. 
CAPACITY  OF  TWO-PHASE  AND  THREE-PHASE  LINES. 

Capacity,  Two-phase.  —  The  charging  current  of  a  single- 
phase  line  was  shown  in  Chapter  XIII  to  be 


*HH) 


amperes  per  mile  of  line 

-i* 

where  b  is  the  tabulated  value  of  capacity  susceptance  per 
mile. 

In  a  two-phase,  four-wire  line,  each  phase  is  quite  similar 
to  a  single-phase  line,  and  so  the  charging  current  per  wire 
is 

\Eb. 

As  this  amount  of  charging  current  flows  in  each  phase, 
the  total  amperes  of  charging  current  are 

Eb, 

for  a  two-phase,  four-wire  line. 

Capacity,  Three-phase,  Irregular  Spacing.  —  When  the 
wires  of  a  three-phase  transmission  line  are  not  spaced  at 
the  corners  of  an  equilateral  triangle,  but  the  transposi- 
tion of  the  conductors  is  carried  out  at  regular  intervals, 
the  charging  current  in  the  wires  will  be  a  balanced,  three- 
phase  current,  since  each  wire  will  have  passed  through  the 
same  average  conditions.  This  is  shown  in  an  approxi- 
mate manner  as  follows: 

93 


94 


TRANSMISSION  LINE  FORMULAS 


As  when  calculating  the  self-induction  of  an  irregularly 

spaced  line,  consider  a  line  three 
miles  long  which  is  transposed 
at  the  end  of  each  mile. 

The  work  in  carrying  a  unit 
charge  from  C  to  B  (Fig.  25) 
assuming  the  charges  concen- 
trated  at  the  centers  of  the 
wires,  is  approximately 


Fig.  25. 


Ea  =  qB  2  loge-  -  qc  2  loge-  +  qA  2  loge-- 

p  p  0 

Now  qB  is  a  periodic  quantity,  which  alternates  in  value 
at  the  same  frequency  as  the  voltage  or  current. 
We  have  qB  =  CiE 

=  _,-v 

27T/ 

where  IB'  is  the  charging  current  flowing  into  the  capacity 
Ci  of  the  wire  B. 
Thus 

/  2  loge  -  -  /</  2  loge  -  +  //  2  loge  £ \ 
p  p  tv 


2  7T/ 


In  the  second  mile  the  conductors  are  transposed  into 
new  positions.  Let  the  currents  in  them  remain  the  same. 
Therefore, 

Ea  =  =l(l  B'  2  loge  -  -  IC'  2  log,  -  +  //  2  log€  -), 

27T\  p  CI 


T/ 

and  in  the  third  mile 

Ea 


~  -  IC'  2 

p 


"  + 


2  loge  -)• 

al 


CAPACITY  OF  TWO-PHASE  AND   THREE-PHASE  LINES        95 


Let 
and 
and 

Then 


Adding  together  and  dividing  by  3,  we  obtain  the  ap- 
proximate average  value  per  mile, 


that  is, 

where 
Similarly, 

and 


Ea=^(V-v)2logA 

5=  ^abc. 
Eb  =  — 4(  V  -  I  A)  2  log€  - 

2  7T/  \  /  p 


Fig.  26. 

Ea=—  i.ooEj  as  in  Fig.  26, 
^b—  (°-5o  —  o.866j)  £, 
Ec=  (o.5o+o.866j)E. 

j  2  irfE  (0.50  +  0.8667) 
5 

P 


(l) 


(0.50  -  0.8667),  (3) 


96  TRANSMISSION  LINE   FORMULAS 

also  I  A  +  IB  +  Ic  =  o,  (4) 

since  they  are  currents  flowing  in  a  three-phase  line. 
From  equations  (i)  and  (3) 

2loge- 

P 
Adding  (4)  and  (5),  we  have 

7    '_        _L_  y  2  irfE  X  1 .00 

3  2loge~ 

P 

-r,        ,,.           T  ,           i    .  .  2  TT/E  (—  0.50  -f  0.866  /) 
From  this,        IB  = j=.  X  — - — • = — • —  • 

3  2 log€ - 

P 

and  I0'—    *x**fE(-°-SQ-<>.*66fl. 

^  2  loge  ~ 

P 

The  vectors  for  IA,  IB  and  /c'  may  now  be  plotted  as 
in  Fig.  27,  and  it  is  seen  that  the  vectors  are  the  same 
,     length  and  are  at  120°  to  each  other. 
Thus  the  charging  current  is  a  bal- 
.  anced  three-phase  current. 

The   power  factor  of  the  charging 


current  is  zero,  since  the  current  in 
any  wire  I  A  (Fig.  27)  is  at  right  angles 
to  the  direction  OP  (Fig.  26)  of   the 
Fig.  27.  corresponding  star  voltage  or  in-phase 

current. 
The  total  amperes  of  charging  current  are 

T    /  27T/E 

I  A  —  — - —  per  centimeter 

2  loge- 
P 


CAPACITY  OF  TWO-PHASE  AND   THREE-PHASE  LINES       97 


,        5 
logio  - 
P 
=  Eb, 

where  b  is  the  tabulated  value  of  capacity  susceptance  per 
mile. 

Capacity,  Three-phase,  Regular  Spacing.  —  When  the 
conductors  are  placed  an  equal  distance,  s,  from  each  other, 
the  formula  for  b  is 

,  38.83  X    27T/Vy  ., 

£  _  o — o j_  x  10  a  mhos  per  mile. 

logio  ~ 
P 

Capacity,  Three-phase,  Flat  Spacing.  —  When  the  wires 
lie  in  one  plane  (either  horizontal  or  vertical),  the  certer 
one  being  at  a  distance  a  from  the  other  two  (see  Fig.  18), 
and  the  wires  being  transposed  at  regular  intervals,  the 
formula  for  susceptance  is 

b=         38.83  X  2  rf    _yjn_. 


,  . 

logio— 
P 

mhos  per  mile  of  transmission  line. 


CHAPTER  XV. 
THEORY  OF   CONVERGENT  SERIES. 

THE  well-known  fundamental  formulas  for  a  transmis- 
sion line  without  branches,  in  which  the  load  is  delivered 
only  at  the  end  of  the  line,  are  as  follows: 

E8  =  E  cosh  VYZ  +  7/b|  /£  sinh  VYZ, 
and          ia  =  /  cosh  VYZ  +~^  sinh  VYZ~ 


where        E8  and  I8  are  the  voltage  and  current  at  the 

supply  end, 
E  and  I  are  the  voltage  and  current  at  the  re- 

ceiver end, 

Y  is  the  line  admittance 
and  Z  is  the  line  impedance. 

The  above  .equations  have  been  published  at  various 
times.     They  are  obtained  as  follows: 

Let  r  =  resistance  of  conductor  per  unit  length, 

and  x  =  reactance  of  conductor  per  unit  length. 

Then  z  =  r  +jx 

=  impedance  of  conductor  per  unit  length. 
Let  g  =  leakage  conductance  from  conductor  per 

unit  length, 
and  b  =  capacity  susceptance  of  conductor  per  unit 

length. 
Then  y  =  g  +jb 

=  admittance  of  conductor  per  unit  length. 
98 


THEORY  OF  CONVERGENT  SERIES         99 

Let  EI  =  voltage  of  line  at  a  distance  /  from  the  re- 

ceiver end, 
and  let        Il  =  Pl-jQl 

=  current  in  the  line  at  a  distance  /  from  the 

receiver  end. 

(Since  Ii  is  usually  not  in  phase  with  the  voltage,  it 
must  be  expressed  as  a  complex  quantity.) 

Now  in  an  element  of  length,  dl,  of  the  line,  the  voltage 
consumed  by  impedance  is 

dEl  =  zltdl. 

The  current  consumed  by  admittance  is 
dlt  =  yEtdl. 

Thus  1f'  =  Z/"  (l) 

and  '  ^  =  yEl:  (2) 

Differentiating  (i) 

d2Et  _    dlj 
dl2       Z  dl  ' 
Substituting  (2)  in  this  gives 


This  is  a  differential  equation  of  the  second  order  and 
may  be  expressed  in  the  form 

(D*  -  yz)  E,  =  o, 
and  we  have 


and  from  (i), 

II  =  z~dT 

(5) 


100  TRANSMISSION  LINE  FORMULAS 

Now  at  the  supply  end, 
yl  =  Y, 

and  zl  =  Z 

Therefore,  Ea  =  A  l€  v™  +  A2e  ~  ^™  (6) 


(7) 

At  the  receiver  end, 

|»o 

and  E  =  A1+A2  (8) 

and  /=Vf  <Xi-^2).  (9) 

From  (6) 


(10) 

which,  by  the  definition  of  cosh  6  and  sinh  6,  is 

(A1  +  A2)  cosh  VYZ  +  (Ai-  At)  sinh 
Therefore,  from  (8)  and  (9), 


E3  =  E  cosh  VYZ  +     =  7Z  sinh  VFZ:        (i  i) 
Similarly,  from  (7), 


=  7  cosh  VFZ  +  —  =  EY  sinh  VFZ.      (13) 

Equations  (n)  and  (13)  are  the  fundamental  formulas  of 
transmission  lines,  as  generally  written. 


THEORY  OF  CONVERGENT  SERIES 


101 


Now 


I     .   YZ  .    F2Z2  \ 

=  21  H 1 hetc.l. 

\          2       2-3 -4  / 

Similarly  cVT1  - 

/=7^  /         FZ  F2Z2  \ 

=  2  v  FZ  i  + H h  etc.  1. 

V        2-3      2-3.4.5  / 

Substituting  these  results  in  (10)  and  (12),  we  can  ex- 
press the  fundamental  equations  as  follows: 


and 

/.   = 


T 

2-3-4 

r2-3-4-5  '6 

r  cue.  i 

FZ 

F2Z2 

F3Z3 

1    r*tr 

r 

-f-etc. 

2-3 

2-3-4- 

5    2-3.4.5-6 

•7 

FZ  , 

F2Z2 

I 

etc.) 

T 

H 
2-3-4 

2.3.4.5-6 

FZ 

F2Z2 

F3Z3 

_  O-Atr    1 

.)  (14) 
/ 


.)*(i5) 


2-3     2.3-4-5     2.3.4.5.6-7 

Equations  (14)  and  (15)  are  the  same  as  those  tabulated 
in  Chapter  VI  for  obtaining  Ea  or  A  +  jB,  and  I8  or  C  +  JD. 
For  no-load  values,  all  that  is  necessary  is  to  put  the  load 
current,  /,  equal  to  zero. 

When  conditions  are  given  at  the  supply  end,  the  same 
equations  for  full-load  conditions  are  obtained,  except  that 

*  See  references  to  T.  R.  Rosebrugh,  J.  F.  H.  Douglas,  and  C.  P.  Stein- 
metz,  page  41. 


102  TRANSMISSION  LINE  FORMULAS 

the  second  half  of  the  expressions  for  voltage  and  current 
is  negative,  since  power  is  now  flowing  away  from  the  point 
where  the  voltage  is  specified,  instead  of  toward  it.  Thus 
we  have  the  series  fox  F  +jG  and  M  +JN. 

At  no  load,  the  conditions  are  really  not  all  specified  at 
the  supply  end,  but  the  current  is  specified  to  be  zero  at 
the  receiver  end,  and  this  necessitates  the  use  of  special 
series.  From  Table  V  we  have  the  ratio  of  the  voltage  at 
the  two  ends  of  the  line, 

EQs  _  AQ  +JB<> 
E  ~         E 

,   YZ  ,     F2Z2 

H 1 hetc 

2        2.3.4 

This  ratio  is  independent  of  the  voltage  E,  and  depends 
only  on  the  constants  of  the  line.  Thus,  if  Es,  the  voltage 
at  the  supply  end  at  no  load,  is  given,  we  can  obtain  the 
no-load  voltage  at  the  receiver  end  from  the  equation, 


etc 


2  2.3-4 

V7        V272  X-1 

or 


2        2-3-4 

which,  when  expanded  by  the  binomial  theorem,  gives 
EQ  = 


-  etc] 

/ 


2  24  720  8064 

(16.) 
as  in  Table  VI. 

The  no-load  current  at  the  supply  end  is 

CV7  1727-2  \ 

i+  —  +-  -+etcl 

2-3      2-3.4.5  / 

as  in  the  equation  for  CQ  +JD0. 


r  '  '       .       • 


THEORY  OF  CONVERGENT  SERIES  103 

Substituting  the  value  of  E0  from  equation  (16),  we  have 
-YZ+^-Y2Z2-  —  Y*Z3+-m-Y*Z*-e 

2  24  720  8064 


2.3     2-3.4.5     2.3.4.5.6.7 

F474  \ 

+  -  ~  -  5—  +  etc.  ). 

2.3.4.5.6.7.8.9  / 

Multiplying  the  two  series  together  by  the  ordinary 
algebraical  method,  we  obtain 


F4Z4- 


3  i5  3i5  2835 

as  given  in  Table  VI  of  convergent  series. 


PART    III 
TABLES. 


TABLE  I.  —  FORMULAS  FOR  SHORT  LINES. 
CONDITIONS  GIVEN  AT  RECEIVER  END. 

These  formulas  are  exact  when  the  line  is  short.     When  the  line  is  20 
miles  long,  they  are  correct  within  approximately  ^  of  i%  of  line  voltage. 
Conditions  given: 
K.V.A.  =  K.V.A.  at  receiver  end. 

E  =  Full  load  voltage  at  receiver  end. 
cos  6  =  Power  factor  at  receiver  end. 
K.W.  =  K.V.A.  cos  0. 

r  =  Resistance  of  conductor  per  mile.  (From  Tables  VII-  VIII.) 
x  =  Reactance  of  conductor  per  mile.  (From  Tables  IX-XII.) 
/  =  Length  of  line  in  miles. 

1000  K.V.A.  cos  8  ,    ,. 

Then      P=  -  =;  -  =  In-phase    current    at    receiver    end    (in 
c, 

total  amps.). 

_       1000  K.V.A.  sin  0      _,  .  ,     ,. 

Q  =  -  —  -  =  Reactive    current    at    receiver    end    (in 

total  amps.)  when  current  is  lagging. 

1000  K.V.A.  sin  0    , 

=  --        —  -  —     —  when  current  is  leading. 
hi 

Find  the  following  quantities: 

Three  phase  or  two  phase.  Single  phase. 

A  =  E  +  Prl+QxL  A  =  E  +  2  Prl  +  2  Qxl. 

B  =  Pxl-  Qrl.  B  =  2Pxl-  2Qrl. 

Formulas  (capacity  neglected)  : 

B*       * 

(1)  Voltage  at  supply  end  =  A  -\  —  -\  • 

R2 

(2)  Regulation  of  line  =  A  +    -    —  E.    (Same  as  line  drop.) 


(3)  Per  cent  regulation  of  line  =  -  -  -  -^  -  '-  per  cent. 

(Same  as  per  cent  line  drop.) 
,.    r          104 


:  '      '    r""  r     '  r  '•      "' 


rrrrffr 


,,,M,|,,,,,,,,",M, ,,,,,,,,,,,,,,,,,, 

in > i* »_ 


T"T"       '"J""!"*1!1 
4k  «M  M 


T-JL,^     M1 

i/ 

P 

]i'         —         ,  , 

.  •  i  .  i  .  1  1  1  1  1  1 
?            Uj 

,,,.,,  ,,,  ,...,  j,  .  ..  i 

ro                — 

.,  1  ,,..  |  ...  .j  ...  i  ,,.,  .j  .,-,.,,. 

cd 

Ohms  Resistano 


r»    ro     j- 
T     L       L 


I    I 

.ro 

les. 


FeetSpacing 

Copper 
Aluminum 


jo.ro     ro 


wNoaxnot^oj  K»  5J  —  FeetSpacing 
-   FeetSpacing     Ml I  MM    I  I    i    r^^r 

T^rnTTTTTT^Alumlnum 
Swo03^01-^04  ^oi  ~"  FeetSpacing 
25  Cycles. 


Q  -<Q 

"  JT^ 


ct §  1  °  | 

mn 


fo- 


o 

Z 


i  Jo  i  i  ii  i  i 


I"".!""'"".!" 

CD  00 


T"1'".!""''"'!1     '     J J 

i0  00  "*l  .  0*  W| 


=  Regulation  Factor. 


pmr 


|*  || 

fe  Ii 
st  ^ 


1 


8 


"\ 

•  ^             ^*5 

f^ 

•^ 

/ 

;  /           .^  ? 

^ 

\ 

y           ii' 

I 

.  ,  .  .  .  IJ  .  .  .  .  1 

1  1  1  J      1 

rrrf  njrrrrrfn 

'1  

1  1  '  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1 

O) 

In 

!^ 

m 

fo               •—               o 

Conductor  per  Mile. 


TABLES  105 


TABLE  I.     (Continued.) 
B2 

(4)  K.V.A.  at  supply  end  = -j^-  X  K.V.A. 

(5)  K.W.  at  supply  end  =  ^  (AP  -  BQ). 

(6)  Power  factor  at  supply  end  —  "" — 


1000.    .     . 


(in  decimals). 
(7)  In-phase  current  at  supply  end  =  ~   in  total  amperes.* 


(8)  Reactive  or  quadrature  current  at  supply  end  = 


in  total  amperes.* 

When  this  quantity  is  positive,  the  current  is  lagging. 
When  this  quantity  is  negative,  the  current  is  leading. 

(9)  K.W.  loss  in  line  =  --  (AP  -  BQ  -  EP). 

(10)  Per  cent  efficiency  of  line  =  -^=  -  ^r  per  cent. 

AJc  —  n\) 

.  Total  amps. 

*  Amperes  per  wire,  three  phase,  =  -  _        • 

V3 

.  .  .  Total  amps. 

Amperes  per  wire,  two  phase,  =  --  *—• 


106  TRANSMISSION  LINE   FORMULAS 

TABLE  II.  —  FORMULAS  FOR  SHORT  LINES. 
CONDITIONS  GIVEN  AT  SUPPLY  END. 

These  formulas  are  exact  when  the  line  is  short.     When  the  line  is  20 
miles  long,  they  are  correct  within  approximately  TV  of  i%  of  line  voltage. 

Conditions  given: 

K.V.A.  =  K.V.A.  at  supply  end. 

E8  =  Full  load  voltage  at  supply  end. 
cos  6  =  Power  factor  at  supply  end. 
K.W.  =  K.V.A.  cos  6. 

r  =  Resistance  of  conductor  per  mile.  (From  Tables  VII-  VIII.) 
x  =  Reactance  of  conductor  per  mile.  (From  Tables  IX-XII.) 
/  =  Length  of  line  in  miles. 

_.        1000  K.V.A.  cos  0      T      . 

Then  JP«  =  -  ~  -  =  In-phase     current    at    supply    end 

/i« 

(hi  total  amps.). 

_       1000  K.V.A.  sin  0 

Qt  =  -  5  -  •  =  Reactive     current     at    supply    end 
•£>« 

(in  total  amperes)  when  current  is 
lagging. 

1000  K.V.A.  sin  9 
=  --     —  =  -  •  when  current  is  leading. 

-c-a 

Find  the  following  quantities: 

Three  Phase  or  Two  Phase.  Single  Phase. 

F  =  E8-  P8rl  -  Q8xl  F  =  E8-2  P8rl  -  2  Qaxl 

G  =  Qarl  -  P8xl  G  =  2  Qarl  -  2  Paxl. 

Formulas  (capacity  neglected): 

(1)  Voltage  at  receiver  end  =  F  -\  —  ^  • 

(2)  Regulation  of  line  =  E8  —  F  --    .     (Same  as  line  drop.) 


(3)  Per  cent  regulation  of  line  =  -  -^  -  -  per  cent. 


(Same  as  per  cent  line  drop.) 


(4)  K.V.A.  at  receiver  end  =  — ^ X  K.V.A. 


TABLES  107 

TABLE  II.     (Continued.) 

(5)  K.W.  at  receiver  end  —  ^^  (FPS  —  GQa). 

i  (FP8  -  GQ8)  Ea_ 

(6)  Power  factor  at  receiver  end  : 


(in  decimals). 
(7)  In-phase  current  at  receiver  end  =  — - — -^- -  in  total  amperes.* 


GPs  +  FQ» 

(8)  Reactive  or  quadrature  current  at  receiver  end  =          — ^  — 

F+TF 

in  total  amperes.* 

When  this  quantity  is  positive,  the  current  is  lagging. 
When  this  quantity  is  negative,  the  current  is  leading. 

(9)  K.W.  loss  in  line  =  •—  (ESPS  -  FPS  +  GQ8). 

(10)  Per  cent  efficiency  of  line  =  — - — £^       —  per  cent. 

Total  amps. 
*  Amperes  per  wire,  three  phase,  = — ?=- 

Total  amps. 
Amperes  per  wire,  two  phase,  = — 


108  TRANSMISSION   LINE    FORMULAS 

TABLE  in.  —  K  FORMULAS  FOR  TRANSMISSION  LINES. 
CONDITIONS  GIVEN  AT  RECEIVER  END. 

Accurate  within  approximately  TV  of  i%  of  line  voltage  up  to  100  miles, 
and  \  of  i%  up  to  200  miles,  for  lines  with  regulation  up  to  20%. 

(cycles)2 
K  =  6  —      '- — .    K=  2.16  for  60  cycles.     K  =  0.375  for  25  cycles. 

Conditions  given: 

K.V.A.  =  K.V.A.  at  receiver  end. 

E  =  Full  load  voltage  at  receiver  end. 
cos  6  =  Power  factor  at  receiver  end. 
K.W.  =  K.V.A.  cos  8. 

r  =  Resistance  of  conductor  per  mile.  (From  Tables  VII-VIII.) 
x  =  Reactance  of  conductor  per  mile.  (From  Tables  IX-XII.) 
/  =  Length  of  transmission  line  in  miles. 

„,,  „       1000  K.V.A.  cos  6 

I  hen  P  =  - =  In-phase    current    at    receiver    end 

Hi 

(in  total  amps.). 

-.       1000  K.V.A.  sin  0      ^ 

Q  =  p; =  Reactive    current    at    receiver    end 

rL 

(in  total  amps.),  when  current  is 
lagging. 

1000  K.V.A.  sin  0     . 
= -= when  current  is  leading. 


TABLES  109 


TABLE  III.     (Continued.} 

Find  the  following  quantities: 

Full  Load. 


€  _  The  above  are  for  two-  and  three-phase  lines.    For  single-phase 
lines  use  2  r  and  2  x  in  place  of  r  and  #. 


110  TRANSMISSION  LINE  FORMULAS 

TABLE  in.     (Continued.) 

CONDITIONS  GIVEN  AT  RECEIVER  END. 
Formulas: 

Full  Load.  No  Load. 

Voltage  at  receiver  end. 


d) 


(for  constant  supply  voltage) 
Regulation  at  receiver  end  in  volts,  for  constant  supply  voltage. 

A+TA 

(3)  -  ng.X-A 


N.B.    The  regulation  at  receiver  end  may  be  expressed  as  a  percentage 
ofE. 

Voltage  at  supply  end. 

UE.-A  +  Z.-  (5)*.-*+*! 

(for  constant  receiver  voltage). 
Regulation  at  supply  end  in  volts,  for  constant  receiver  voltage. 

,,.     *   ,    &        A         B<? 
(6)   A  H  --  j  —  AQ  --  j-  • 

2  A  2  AQ 

N.B.    The  regulation  at  supply  end  may  be  expressed  as  a  percentage 


TABLES  HI 


TABLE  III.     (Continued.') 
Current  at  supply  end  in  total  amperes.* 

(7)  VcnTo2.  (8) 

K.V.A.  at  supply  end. 


fiT.PF.  at  supply  end. 

(n)  (4C  +  5Z>).  (12) 


Power  factor  at  supply  end,  in  decimals. 

AC  +  BD  ,    ,  A0C0+B0D0 

(13) 


In-phase  current  at  supply  end  in  total  amperes* 
AC+BD  ,  , 


Reactive  current  at  supply  end  in  total  amperes* 
,    .     BC-AD  ,  Os     B0 

(I7)       "'  ( 


2 

When  this  quantity  is  positive,  the  current  is  lagging. 
When  this  quantity  is  negative,  the  current  is  leading. 
K.W.  loss  in  line. 
(19)    —  (AC  +  BD  -EP).  (20)  — 

IOOO  IOOO 

[same  as  No.  12]. 
Per  cent  efficiency  of  line. 
.       100  EP 

percent' 


Total  amps. 
*  Amperes  per  wire,  three-phase,  =  -  -=  -  • 

^3 

Total  amps. 
Amperes  per  wire,  two  phase,  =  -  p—  • 


112  TRANSMISSION  LINE  FORMULAS 

TABLE  IV.  —  K  FORMULAS  FOR  TRANSMISSION  LINES. 
CONDITIONS  GIVEN  AT  SUPPLY  END. 

Accurate  within  approximately  fa  of  i%  of  line  voltage  up  to  100  miles 
and  |  of  i%  up  to  200  miles,  for  lines  with  regulation  up  to  20%. 

K  =  —  —  -  —  .    K  =  2.16  for  60  cycles.    K  =  0.375  for  25  cycles. 
10,000 

Conditions  given: 

K.V.A.  =  K.V.A.  at  supply  end. 

Es  =  Full  load  voltage  at  supply  end. 
cos  6  =  Power  factor  at  supply  end. 
K.W.  =  K.V.A.  cos0. 

r  =  Resistance  of  conductor  per  mile.  (From  Tables  VII-VIII.) 
x  =  Reactance  of  conductor  per  mile.  (From  Tables  IX-XII.) 
I  =  Length  of  transmission  line  in  miles. 

Then  P8  =  I0°°  K'Y'A'  C°S  °  =  In-phase  current  at  supply  end  (in 

&8 

total  amps.). 

t  supply  end  (in 


total  amps.),  when  current  is  lagging. 

looo  K.V.A.  sin  9     ,  ,  .    .     ,. 

=  -  when  current  is  leading. 


TABLES 


TABLE  IV.     (Continued.) 
Find  the  following  quantities: 

Full  Load. 


iooo 


Kf     I 


No  Load. 

(— Ti 

Viooo/  ) 


Note.  —  The  above  are  for  two-  and  three-phase  lines.    For  single- 
phase  lines  use  2  r  and  2  *  in  place  of  r  and  £. 


114  TRANSMISSION  LINE  FORMULAS 

TABLE  IV.     (Continued.} 

CONDITIONS  GIVEN  AT  SUPPLY  END. 
Formulas: 

Full  Load.  No  Load. 

Voltage  at  receiver  end. 


(for  constant  supply  voltage). 

Regulation  at  receiver  end  in  volts,  for  constant  supply  voltage. 
Go2  G2 


N.B.     The  regulation  at  receiver  end  may  be  expressed  as  a  percentage 
of  E. 

Voltage  at  supply  end. 


(for  constant  receiver  voltage). 
Regulation  at  supply  end,  in  volts,  for  constant  receiver  voltage. 

F  +  121 
(6)  Es  -  -  -  if-  E.. 


N.B.    The  regulation  at  supply  end  may  be  expressed  as  a  percentage 
of  E,. 


TABLES 

TABLE  IV.     (Continued.) 
Current  in  total  amperes* 


(7)   VM2  +  N*  at  receiver  end.  (8)    ^M<?  +  N<?  at  supply  end.- 

K.V.A. 
(Q\   _*_  IP+GL]  VM2  +  #2  at  receiver  end. 

7     1000  \  2V  I 

(I0)   _JL_  £a  VM  02  +  No2  at  supply  end. 

' 


1000 


(„) 


^-(^M  +  G^)  at  receiver  end.  (12)  £8M0  at  supply  end. 


IOOO 


(13)  /  rm^:GN    - at receiver end- 


Power  factor,  in  decimals. 

—  at  receiver 

(14)  at  supply  end. 

VM<?  +  No2 

In-phase  current  in  total  amperes* 

(l,\   FM  +  GN  at  recdver  end>  (l6)  MQ  at  supply  end. 


Reactive  current  in  total  amperes* 

(I7)   GM-FN  at  recdver  end  (Ig)  jv0  at  supply  end. 


When  this  quantity  is  positive,  the  current  is  lagging. 

When  this  quantity  is  negative,  the  current  is  leading. 

K.W.  loss  in  line. 

(IQ)    -±-(E8P8-FM-GN).  (20)  ^£sM0(sameasNo.i2). 

Per  cent  efficiency  of  line. 

(21)   =r-p —    —  per  cent. 

Total  amps. 
*  Amperes  per  wire,  three  phase,  =  -  — — — 

O 

Total  amps. 
Amperes  per  wire,  two  phase,  =        — - — 


Il6  TRANSMISSION  LINE  FORMULAS 

TABLE  V.  —  CONVERGENT  SERIES  FOR  TRANSMISSION  LINES. 
CONDITIONS  GIVEN  AT  RECEIVER  END. 

The  convergent  series  give  the  results  of  the  fundamental  formulas  as 
accurately  as  desired,  if  a  sufficient  number  of  terms  is  used. 

When  conditions  are  given  at  the  receiver  end,  the  same  as  with  the  K 
formulas,  find  the  quantities: 

Full  Load. 


2-3-4-S       2-3.4.5.6.7 


2.3     2.3.4.5     2.3.4.5.. 

where  Z  =  (r  +./*)*• 

r  =  resistance  of  conductor  per  mile. 
x  =  reactance  of  conductor  per  mile. 
I  =  length  of  transmission  line  in  miles. 


+  etc. 


g  =  leakage  conductance  of  conductor  per  mile. 
b  =  capacity  susceptance  of  conductor  per  mile. 

Use  A,  B,  C,  D,  etc.,  with  the  equations  in  the  third  and  fourth  pages 
of  Table  III  to  solve  transmission  line  problems. 

£2  _ 

Note  i.  —  In  the  formulas,   A  +  —  7-  is  used  instead  of  V  A2  -f-  .B2. 

zA 

This  approximation  may  be  used  for  very  accurate  work,  as  it  is  correct 
within  approximately  T§^  of  i%  when  the  regulation  is  not  more  than  20%. 
Note  2.  —  The  above  are  for  two-  and  three-phase  lines.  For  single- 
phase  lines  use  2  r  and  2  x  in  place  of  r  and  x,  and  use  \  g  and  %  b  in  place 
of  g  and  b. 


TABLES  117 

TABLE  VI.  —CONVERGENT  SERIES  FOR  TRANSMISSION  LINES. 

CONDITIONS  GIVEN  AT  SUPPLY  END. 

The  convergent  series  give  the  results  of  the  fundamental  formulas  as 
accurately  as  desired,  if  a  sufficient  number  of  terms  is  used. 

When  conditions  are  given  at  the  supply  end,  the  same  as  with  the 
K  formulas,  find  the  quantities: 

Full  Load. 

Z3  etc 


2-3-4       2-3-  4-5- 

—  +       F2Z2      +  -  F3Z3  A       +  etc 
2-  3^2-  3-  4.  5      2.3.4.5.6-7 


„{     ,    FZ    ,        F2Z2  PZ3  u    \ 

—  E3Y   i  4  ---  ---  --  -  --  r-  etc.  I  . 

V      -  2-3       2-3.4.5       2-3-4-S-6-7  / 

No  Load. 
iYZ+  ^-Y*Z*-  —  F3Z3  +  ^  F4Z4  -  etc.Y 

24  720     _      8004  y 


-  I  FZ  -h  —  F2Z2  -  --  F3Z3  + 


15          3js  2835 

where  Z  =  (r  +jx)  I. 

r  =  resistance  of  conductor  per  mile. 
x  =  reactance  of  conductor  per  mile. 
I  =  length  of  transmission  line  in  miles. 
Y=(g+jb)l. 

g  =  leakage  conductance  of  conductor  per  mile. 
b  =  capacity  susceptance  of  conductor  per  mile. 

Use  F,  G,  M,  N,  etc.,  with  the  equations  in  the  third  and  fourth  pages 
of  Table  IV  to  solve  transmission  line  problems. 

G2  /  - 

Note  i.  —  In  the  formulas,  F  -\  --  -  is  used  instead  of  V  F2  +  G3.    This 

2r 

approximation  may  be  used  for  very  accurate  work,  as  it  is  correct  within 
approximately  Tf^  of  *%  when  the  regulation  is  not  more  than  20%. 

Note  2.  —  The  above  are  for  two-  and  three-phase  lines.  For  single- 
phase  lines  use  2  r  and  2  x  in  place  of  r  and  x,  and  use  |  g  and  £  6  in  place  of 
g  and  b. 


n8 


TRANSMISSION  LINE   FORMULAS 


TABLE  VII.  — RESISTANCE  OF  COPPER  WIRE  AND  CABLE. 

Data  assumed: 

Temperature,  20°  C.  (68°  F.). 

Conductivity  of  hard  drawn  copper,  97.3%  of  the  annealed  copper 

standard. 
Increase  of  resistance  of  cables  due  to  spiralling,  i%. 

Copper  Wire. 


Diam- 

Resistance,  ohms  per  mile. 

B.  &  S. 

Circular 

ctcr 

gauge. 

mils. 

(2P) 

inches. 

Direct 
current. 

25  cycles. 

In- 
crease. 

60 

cycles. 

In- 
crease. 

Per  cent 

Per  cent 

0000 

2ll,6oo 

.4600 

.  .  . 

.2655 

.2657 

.08 

.2667 

•44 

000 

167,800 

.4096 

•3348 

•3350 

•05 

.3358 

.27 

00 

133,100 

.3648 

.4221 

.4223 

•03 

.4229 

•I? 

o 

105,500 

•3249 

•5326 

.5327 

.02 

•5331 

.  II 

I 

83,690 

.2893 

... 

.6714 

.6714 

.OI 

.6718 

.07 

2 

66,370 

.2576 

.8466 

.8466 

.OI 

.8469 

.04 

3 

52,630 

.2294 

1.068 

1.068 

1.  068 

•03 

4 

41,740 

.2043 

1.346 

1.346 

I.346 

.02 

5 

33,ioo 

.1819 

1.697 

1.697 

1.698 

.01 

6 

26,250 

.1620 

2.  I4O 

2.  I4O 

2.140 

.OI 

7 

20,820 

•1443 

2.699 

2.699 

2.699 

8 

16,510 

.1285 

3-403 

3-403 

3.403 

TABLES 


119 


•TABLE  VII.     (Continued.) 
Copper  Cable. 


Diam- 

No. of 

Resistance,  ohms  per  mile. 

B.  &  S. 

gauge. 

Circular 
mils. 

eter 

(2p) 

inches. 

wires 
as- 
sumed. 

Direct 
current. 

25  cycles. 

In- 
crease. 

60 

cycles. 

In- 
crease. 

500,000 

.8lII 

19 

•1135 

.1140 

Per  cent 
•41 

.1162 

Per  cent 
2-34 

.... 

450,000 

.7695 

19 

.1261 

.1265 

•33 

.1285 

1.90 

.... 

400,000 

.7255 

19 

.1419 

.1422 

.26 

.1440 

1.50 

350,000 

.6786 

J9 

.1621 

.1625 

.20 

.1640 

1.16 

300,000 

.6211 

7 

.1892 

.1894 

•15 

.1908 

•85 

250,000 

.5669 

7 

.2270 

.2272 

.10 

.2284 

.60 

oooo 

211,600 

.5216 

7 

.2682 

.2684 

.07 

.2693 

•43 

ooo 

167,800 

•4645 

7 

.3382 

.3384 

•05 

•3391 

•27 

oo 

133,100 

•4137 

7 

.4264 

•4265 

•03 

.4271 

•17 

0 

105,500 

•3683 

7 

•5379 

.5380 

.02 

•5385 

.11 

I 

83,690 

.3280 

7 

.6781 

.6782 

.01 

.6785 

.07 

2 

66,370 

.2921 

7 

•8550 

•8551 

.01 

.8554 

.04 

3 

52,630 

.2601 

7 

1.078 

1.078 

1.078 

-.03 

4 

41,740 

.2317 

7 

1.360 

1.360 

1.360  . 

.02 

TABLE  VII.     (Continued.}—  TEMPERATURE   COEFFICIENTS  OF 

COPPER. 
For  different  initial  temperatures  and  different  conductivities. 


Ohms  per 
meter-gram 
at  20  deg. 
cent. 

Per  cent 
conduc- 
tivity. 

«0 

«15 

"20 

«25 

"30 

«50 

.16108 

95 

.00405 

.00381 

.00374 

.00367 

.00361 

.00336 

•15940 

96 

.  00409 

.00386 

.00378 

.00371 

.00364 

.00340 

•15776 

97 

.00414 

.00390 

.00382 

•00375 

00368 

•00343 

.15727 

97-3 

.00415 

.00391 

.00383 

.00376 

.00369 

.00344 

.15614 

98 

.00418 

.00394 

.00386 

.00379 

.00372 

.00346 

•15457 

99 

.00423 

.00398 

.00390 

.00383 

•00375 

.00349 

.153022 

TOO 

.00428 

.OO4O2 

•00394 

.00386 

•00379 

•00352 

•I5I5I 

IOI 

.00432 

.OO4O6 

.00398 

.00390 

.00383 

•00355 

where  Rt  is  the  resistance  at  any  temperature  t  deg.  cent. 

and  Rti  is  the  resistance  at  any  "initial  temperature"  ti  deg.  cent. 

From  Appendix  E,  Standardization  Rules  of  the  A.I.  E.  E.,  June  27, 1911. 


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TABLE  XVII.  — POWER   FACTOR  TABLE. 


Cose. 

Sin  e  (approx.). 

Sine. 

•95 

.312 

.312250 

.90 

•436 

.435890 

•85 

.527 

.526783 

.80 

.600 

.600000 

•75 

.661 

.661438 

.70 

.714 

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•65 

.760 

•759934 

.60 

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•55 

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•So 

.866 

.866025 

INDEX 

PAGE 

Absolute  value  of  a  complex  quantity 57 

Admittance 98 

Aluminum  cables,  Table  VIII 120 

Annealed  copper  standard 55 

Capacity 61 

Capacity,  derivation  of  formulas: 

Cables 89 

Effect  of  the  earth 90 

Two  round  wires 84 

Two-phase 93 

Three-phase,  irregular  spacing 93 

Three-phase,  regular  flat  spacing 97 

Three-phase,  regular  spacing 97 

Capacity  susceptance 6 

of  cable,  25  cycles,  Table  XIV 128 

of  cable,  60  cycles,  Table  XVI 131 

of  wire,  25  cycles,  Table  XIII 127 

of  wire,  60  cycles,  Table  XV 130 

Charging  current 6 

Chart,  regulation,  description  of 8 

Complex  numbers 42 

Condenser  effect 6 

Conductance 43 

Conductivity  of  copper 55 

of  aluminum 55 

Conductors 53 

Convergent  series 45,  1 16 

derivation  of 98 

description  of 41 

Corona 6 

Diameters  of  wires  and  cables 118 

Elements  of  a  transmission  line 4 


136  INDEX 

PAGE 

Fundamental  transmission  line  formulas 41,  98 

Hyperbolic  transmission  line  formulas 41,  98 

Image  wire 90 

Impedance 98 

"  J"  terms 42 


"K"  formulas 28,  108 

"K "  formulas,  description  of 25 


Leakage  conductance 43 

Leakage  current 6 

Line  drop 9 

Loss  in  line,  derivation  of  formula 61 

Magnetic  field  around  wire 5,  62 

Power  factor  table 133 

Problem,  5000  feet 14,  23 

3  miles,  80%  P.F 14,  23 

3  miles,  85%  P.F 12 

10  miles 22 

15  miles 13,  22 

20  miles 24 

25  miles 14,  24 

75  miles 13,  15,  3» 

80  miles 12,  50 

100  miles,  60,000  volts 39,  51 

100  miles,  66,000  volts n,  36,  38,  50 

100  miles,  1 10,000  volts 15 

155.34  miles 39>  5° 

200  miles 38,  47 

300  miles 39*5! 

300  miles,  with  substation 36,  48 

400  miles,  with  substation 4°.  52 

Quadrature  volt-amperes,  derivation  of  formula 60 


INDEX  137 

.PAGE 

Reactance  of  cable,  25  cycles,  Table  X 122 

of  cable,  60  cycles,  Table  XII 12$ 

of  wire,  25  cycles,  Table  IX 121 

of  wire,  60  cycles,  Table  XI 124 

Reactance,  derivation  of  formulas: 

single-phase,  wire,  effect  of  flux  in  air 62 

single-phase,  wire,  effect  of  flux  in  the  conductor 64 

single-phase,  cable 76 

two-phase 79 

three-phase,  irregular  spacing 79 

three-phase,  regular  flat  spacing 83 

three-phase,  regular  spacing 83 

Reactance  drop $ 

Reactive  power,  derivation  of  formula 60 

Regulation -. 8,  58 

Regulation  chart,  description  of 8 

Resistance  of  aluminum  cable,  Table  VIII 120 

of  copper  wire  and  cable,  Table  VII 118 

Resistance  drop 4 

Series,  convergent 45,  116 

derivation  of 98 

description  of 41 

Short  line  formulas 18,  104 

description  of 16 

Skin  effect .  •  4 

Skin  effect,  derivation  of  formula 67 

Skin  effect  formula,  proof  by  infinite  series 71 

Spacing 4,  10 

equivalent 10 

flat 10 

two-phase 10 

Substation  loads 26 

Susceptance  (see  also  "  Capacity  ") 6 

Temperature  coefficients 119,  120 

description  of 55 

Watts,  derivation  of  formula 59 

Wire  and  cable  tables n§ 


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Alternating-  Current 
Machines 

Being  the  Second  Volume  of 

Dynamo  Electric  Machinery 

Its  Construction,  Design 
and  Operation 

By 

Samuel  Sheldon,  A.M.,  Ph.D.,  D.Sc, 

Prof,  of  Physics  and  Elec.  Engineering,  Polytechnic  Inst.,  Brooklyn 

AND 

Hobart  Mason,  B.S.,E.E. 

AND 

Erich  Hausmann,  B.S.,-E.E. 

THIS  BOOK  has  been  entirely  re-written  to  bring  it  into  accord 
with  the  present  conditions  of  practice.  It  contains  one-third  more 
material  than  the  first  edition,  two-thirds  of  which  is  new  matter. 
It  is  intended  for  use  as  a  text-book  in  technical  schools  and  colleges  for 
students  pursuing  either  electrical  or  non-electrical  engineering  courses. 
It  is  so  arranged  that  portions  may  be  readily  omitted  in  the  case  of  the 
latter  without  affecting  the  correlation  of  the  remaining  parts.  The  new 
matter  contains  numerous  problems,  rigid  derivations  of  the  fundamental 
laws  of  alternating  current  circuits,  vector  diagrams  pertaining  to  the  cir- 
cuits of  alternating  current  machines,  methods  of  determining  from  tests 
the  behavior  of  machines  and  of  predetermining  their  behaviour  from 
design  data.  There  is  also  given  a  method  of  calculating  all  the  leakage 
fluxes  of  the  induction  motor.  The  last  chapter  gives  complete  directions 
for  the  design  and  construction  of  transmission  lines  in  accordance  with 
modern  practice. 

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Electrical  Engineer's  Pocketbook 

The  Most  Complete  Book  of  Its  Kind  Ever  Published, 

Treating  of  the  Latest  and  Best  Practice 

in  Electrical  Engineering 

By  Horatio  A.  Foster 

Member  Am.  Inst.  E.  E.,  Member  Am.  Soc.  M.  E. 

With  the  Collaboration  of  Eminent  Specialists 

CONTENTS 

Symbols,  Units,  Instruments 
Measurements 
Magnetic  Properties  of  Iron 
Electro  Magnets 
Properties  of  Conductors 
Relations  and  Dimensions  of  Con- 
ductors 

Underground  Conduit  Construction 
Standard  Symbols 
Cable  Testing 
Dynamos  and  Motors 
Tests  of  Dynamos  and  Motors 
.  The  Static  Transformer 
Standardization  Rules 
Illuminating  Engineering 
Electric  lighting  (Arc) 
Electric  lighting  (Incandescent) 
Electric  Street  Railways 
Electrolysis 
Transmission  of  Power 
Storage  Batteries 
Switchboards 
I/ghtning  Arresters 
Electricity  Meters 
Wireless  Telegraphy 
Telegraphy 
Telephony 

Electricity  in  the  U.  S.  Army 
Electricity  in  the  U.  S.  Navy 
Resonance 

Electric  Automobiles          X-Rays  Lightning  Conductors 

Electro-chemistry  and       Electric  Heating,  Cooking       Mechanical  Section 
Electro-metallurgy and  Welding  Index 

D.  VAN    NOSTRAND    COMPANY 

23  MURRAY  and  27  WARREN  STREETS         NEW  YORK 


, 

OVERDUE.  $I'°°    ON    THE 


C, 


•YB  2419 


^Engineering 
Library 

UNIVERSITY  OF  CALIFORNIA  LIBRARY 


